Category: Big Ideas Math

Big Ideas Math Book Geometry Answer Key Chapter 9 Right Triangles and Trigonometry

Right Triangles and Trigonometry Maintaining Mathematical Proficiency

Simplify the expression.

Question 1.
√75

Answer:
square root of 75 = 5,625.

Explanation:
In the above-given question,
given that,
√75.
square root of 75 = 75 x 75.
75 x 75 = 5,625.
√75 = 5,625.

Question 2.
√270

Answer:
square root of 270 = 72,900.

Explanation:
In the above-given question,
given that,
√270.
square root of 270 = 270 x 270.
270 x 270 = 72900.
√270 = 72900.

Question 3.
√135

Answer:
square root of 135 = 18225.

Explanation:
In the above-given question,
given that,
√135.
square root of 135 = 135 x 135.
135 x 135 = 18225.
√135 = 18,225.

Question 4.
\(\frac{2}{\sqrt{7}}\)

Answer:
2/49 = 0.04.

Explanation:
In the above-given question,
given that,
square root of 7 = 7 x 7.
7 x 7 = 49.
\(\frac{2}{\sqrt{7}}\).
2/49 = 0.04.

Question 5.
\(\frac{5}{\sqrt{2}}\)

Answer:
5/4 = 1.25.

Explanation:
In the above-given question,
given that,
square root of 2 = 2 x 2.
2 x 2 = 4.
\(\frac{5}{\sqrt{2}}\).
5/4 = 1.25.

Question 6.
\(\frac{12}{\sqrt{6}}\)

Answer:
12/36 = 0.33.

Explanation:
In the above-given question,
given that,
square root of 6 = 6 x 6.
6 x 6 = 36.
\(\frac{12}{\sqrt{6}}\).
12/36 = 0.33.

Solve the proportion.

Question 7.
\(\frac{x}{12}=\frac{3}{4}\)

Answer:
x = 9.

Explanation:
In the above-given question,
given that,
\(\frac{x}{12}=\frac{3}{4}\)
x/12 = 3/4.
4x = 12 x 3.
4x = 36.
x = 36/4.
x = 9.

Question 8.
\(\frac{x}{3}=\frac{5}{2}\)

Answer:
x = 7.5.

Explanation:
In the above-given question,
given that,
\(\frac{x}{3}=\frac{5}{2}\)
x/3 = 5/2.
2x = 5 x 3.
2x = 15.
x = 15/2.
x = 7.5.

Question 9.
\(\frac{4}{x}=\frac{7}{56}\)

Answer:
x = 32.

Explanation:
In the above-given question,
given that,
\(\frac{4}{x}=\frac{7}{56}\)
4/x = 7/56.
7x = 56 x 4.
7x = 224.
x = 224/7.
x = 32.

Question 10.
\(\frac{10}{23}=\frac{4}{x}\)

Answer:
x = 9.2.

Explanation:
In the above-given question,
given that,
\(\frac{10}{23}=\frac{4}{x}\)
x/4 = 10/23.
10x = 23 x 4.
10x = 92.
x = 92/10.
x = 9.2.

Question 11.
\(\frac{x+1}{2}=\frac{21}{14}\)

Answer:
x = 135.

Explanation:
In the above-given question,
given that,
\(\frac{x + 1}{2}=\frac{21}{14}\)
x+12 x 2 = 21×14.
2x + 24= 294 .
2x = 294 – 24.
2x = 270.
x = 270/2.
x = 135.

Question 12.
\(\frac{9}{3 x-15}=\frac{3}{12}\)

Answer:
x = 6.33.

Explanation:
In the above-given question,
given that,
\(\frac{9}{3 x-15}=\frac{3}{12}\)
27x – 135 = 3×12.
27x = 36 + 135.
27x = 171.
x = 171/27.
x = 6.33.

Question 13.
ABSTRACT REASONING
The Product Property of Square Roots allows you to simplify the square root of a product. Are you able to simplify the square root of a sum? of a diffrence? Explain.

Answer:
Yes, I am able to simplify the square root of a sum.

Explanation:
In the above-given question,
given that,
The product property of square roots allows you to simplify the square root of a product.
√3 + 1 = √4.
√4 = 4 x 4.
16.
√3 – 1 = √2.
√2 = 2 x 2.
4.

Right Triangles and Trigonometry Mathematical practices

Monitoring progress

Question 1.
Use dynamic geometry software to construct a right triangle with acute angle measures of 30° and 60° in standard position. What are the exact coordinates of its vertices?

Answer:

Question 2.
Use dynamic geometry software to construct a right triangle with acute angle measures of 20° and 70° in standard position. What are the approximate coordinates of its vertices?
Answer:

9.1 The Pythagorean Theorem

Exploration 1

Proving the Pythagorean Theorem without Words

Work with a partner.

a. Draw and cut out a right triangle with legs a and b, and hypotenuse c.

Answer:

Explanation:
In the above-given question,
given that,
proving the Pythagorean theorem without words.
a² + b² = c^2.

b. Make three copies of your right triangle. Arrange all tour triangles to form a large square, as shown.

Answer:
a² + b² = c^2.

Explanation:
In the above-given question,
given that,
make three copies of your right triangle.
a² + b² = c^2.

c. Find the area of the large square in terms of a, b, and c by summing the areas of the triangles and the small square.

Answer:
The area of the large square = a² x b^2.

Explanation:
In the above-given question,
given that,
the area of the square = l x b.
where l = length, and b = breadth.
the area of the square = a x b.
area = a² x b^2.

d. Copy the large square. Divide it into two smaller squares and two equally-sized rectangles, as shown.

Answer:

e. Find the area of the large square in terms of a and b by summing the areas of the rectangles and the smaller squares.
Answer:

f. Compare your answers to parts (c) and (e). Explain how this proves the Pythagorean Theorem.

Answer:
a² + b² = c^2.

Explanation:
In the above-given question,
given that,
The length of the a and b is equal to the hypotenuse.
a² + b² = c^{2.
where a = one side and b = one side.}

Exploration 2

Proving the Pythagorean Theorem

Work with a partner:

a. Draw a right triangle with legs a and b, and hypotenuse c, as shown. Draw the altitude from C to \(\overline{A B}\) Label the lengths, as shown.

Answer:
a² + b² = c^2.

Explanation:
In the above-given question,
given that,
a² + b² = c^2.

b. Explain why ∆ABC, ∆ACD, and ∆CBD are similar.

Answer:
In a right-angle triangle all the angle are equal.

Explanation:
In the above-given question,
given that,
∆ABC, ∆ACD, and ∆CBD are similar.
In a right-angle triangle all the angles are equal.

REASONING ABSTRACTLY
To be proficient in math, you need to know and flexibly use different properties of operations and objects.
Answer:

c. Write a two-column proof using the similar triangles in part (b) to prove that a² + b² = c²

Answer:
a² + b² = c²

Explanation:
In the above-given question,
given that,
In the pythagorean theorem.
a² + b² = c^{2
the length of the hypotenuse ie equal to the two side lengths.}
a² + b² = c²

Communicate Your Answer

Question 3.
How can you prove the Pythagorean Theorem?

Answer:
a² + b² = c²

Explanation:
In the above-given question,
given that,
In the pythagorean theorem,
the length of the hypotenuse is equal to the length of the other two sides.
hypotenuse = c.
length = a.
the breadth = b.
a² + b² = c²

Question 4.
Use the Internet or sonic other resource to find a way to prove the Pythagorean Theorem that is different from Explorations 1 and 2.
Answer:

Lesson 9.1 The Pythagorean Theorem

Monitoring Progress

Find the value of x. Then tell whether the side lengths form a Pythagorean triple.

Question 1.

Answer:
x = √52.

Explanation:
In the above-given question,
given that,
the side lengths are 6 and 4.
a² + b² = c^{2
6 x 6 + 4 x 4 = c2
36 + 16 = c2
52 = c2.
c = √52.}

Question 2.

Answer:
x = 4.

Explanation:
In the above-given question,
given that,
the side lengths are 3 and 5.
^{x2 + 3 x 3 = 5 x 5.
x2 + 9 = 25.
x2 = 25 – 9.
x2 = 16.}
x = 4.

Question 3.
An anemometer is a device used to measure wind speed. The anemometer shown is attached to the top of a pole. Support wires are attached to the pole 5 feet above the ground. Each support wire is 6 feet long. How far from the base of the pole is each wire attached to the ground?

Answer:
x = √11.

Explanation:
In the above-given question,
given that,
An anemometer is a device used to measure wind speed.
support wires are attached to the pole 5 feet above the ground.
Each support wire is 6 feet long.
^{d2 + 6  x 6 = 5 x 5.
d2 + 36 = 25.
d2 = 25 – 36.
d2 = 11.
d} = √11.

Tell whether the triangle is a right triangle.

Question 4.

Answer:
Yes the triangle is a right triangle.

Explanation:
In the above-given question,
given that,
the hypotenuse = 3 √34.
one side = 15.
the other side = 9.
so the triangle is a right triangle.

Question 5.

Answer:
Yes, the triangle is a actute triangle.

Explanation:
In the above-given question,
given that,
the hypotenuse = 22.
one side = 26.
the other side = 14.
so the triangle is a acute triangle.

Question 6.
Verify that segments with lengths of 3, 4, and 6 form a triangle. Is the triangle acute, right, or obtuse?

Answer:
Yes the lengths of the triangle form a acute triangle.

Explanation:
In the above-given question,
given that,
the side lenths of 3, 4, and 6 form a triangle.
6 x 6 = 3 x 3 + 4 x 4.
36 = 9 + 16.
36 = 25.
so the length of the triangle forms a acute triangle.

Question 7.
Verify that segments with lengths of 2, 1, 2, 8, and 3.5 form a triangle. Is the triangle acute, right, or obtuse?
Answer:

Exercise 9.1 The Pythagorean Theorem

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
What is a Pythagorean triple?
Answer:

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.

Find the length of the longest side.

Answer:
The length of the longest side = 5.

Explanation:
In the above-given question,
given that,
the side lengths are 3 and 4.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 4 x 4 + 3 x 3.
X x X = 16 + 9.
X x X = 25.
X = 5.

Find the length of the hypotenuse

Answer:
The length of the hypotenuse = 5.

Explanation:
In the above-given question,
given that,
the side lengths are 3 and 4.
in the Pythagorean theorem,
the longest side is equal to the side lengths.
X = 4 x 4 + 3 x 3.
X x X = 16 + 9.
X x X = 25.
X = 5.

Find the length of the longest leg.

Answer:
The length of the longest leg = 5.

Explanation:
In the above-given question,
given that,
the side lengths are 3 and 4.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 4 x 4 + 3 x 3.
X x X = 16 + 9.
X x X = 25.
X = 5.

Find the length of the side opposite the right angle.

Answer:
The length of the side opposite to the right angle = 5.

Explanation:
In the above-given question,
given that,
the side lengths are 3 and 4.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 4 x 4 + 3 x 3.
X x X = 16 + 9.
X x X = 25.
X = 5.

Monitoring progress and Modeling with Mathematics

In Exercises 3-6, find the value of x. Then tell whether the side lengths form a Pythagorean triple.

Question 3.

Answer:

Question 4.

Answer:
The length of the x = 34.

Explanation:
In the above-given question,
given that,
the side lengths are 30 and 16.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 16 x 16 + 30 x 30.
X x X = 256 + 900.
X x X = 1156.
X = 34.

Question 5.

Answer:

Question 6.

Answer:
The length of the x = 7.2.

Explanation:
In the above-given question,
given that,
the side lengths are 6 and 4.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 4 x 4 + 6 x 6.
X x X = 16 + 36.
X x X = 52.
X = 7.2.

In Exercises 7 – 10, find the value of x. Then tell whether the side lengths form a Pythagorean triple.

Question 7.

Answer:

Question 8.

Answer:
The length of the X = 25.6.

Explanation:
In the above-given question,
given that,
the side lengths are 24 and 9.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 24 x 24 + 9 x 9.
X x X = 576 + 81.
X x X = 657.
X = 25.6.

Question 9.

Answer:

Question 10.

Answer:
The length of the x = 11.4.

Explanation:
In the above-given question,
given that,
the side lengths are 7 and 9.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 7 x 7 + 9 x 9.
X x X = 49 + 81.
X x X = 130.
X = 11.4.

ERROR ANALYSIS
In Exercises 11 and 12, describe and correct the error in using the Pythagorean Theorem (Theorem 9.1).

Question 11.

Answer:

Question 12.

Answer:
x = 24.

Explanation:
In the above-given question,
given that,
the side lengths are 26 and 10.
26 x 26 = a x a + 10 x 10.
676 = a x a + 100.
676 – 100 = a x a.
576 = a x a.
a = 24.
x = 24.

Question 13.
MODELING WITH MATHEMATICS
The fire escape forms a right triangle, as shown. Use the Pythagorean Theorem (Theorem 9. 1) to approximate the distance between the two platforms. (See Example 3.)

Answer:

Question 14.
MODELING WITH MATHEMATICS
The backboard of the basketball hoop forms a right triangle with the supporting rods, as shown. Use the Pythagorean Theorem (Theorem 9.1) to approximate the distance between the rods where the meet the backboard.

Answer:
x = 9.1.

Explanation:
In the above-given question,
given that,
the side lengths are 13.4 and 9.8.
13.4 x 13.4 = X x X + 9.8 x 9.8.
179.56 = X x X + 96.04.
179.56 – 96.04 = X x X.
83.52 = X x X.
X = 9.1.
In Exercises 15 – 20, tell whether the triangle is a right triangle.

Question 15.

Answer:

Question 16.

Answer:
No, the triangle is not a right triangle.

Explanation:
In the above-given question,
given that,
the side lengths are 23 and 11.4.
the hypotenuse = 21.2.
21.2 x 21.2 = 23 x 23 + 11.4 x 11.4.
449.44 = 529 + 129.96.
449.44 = 658.96.
449 is not equal to 658.96.
so the triangle is not a right triangle.

Question 17.

Answer:

Question 18.

Answer:
No, the triangle is not a right triangle.

Explanation:
In the above-given question,
given that,
the side lengths are 5 and 1.
the hypotenuse = √26.
26 x 26 = 5 x 5 + 1 x 1.
676 = 25 + 1.
676 = 26.
676 is not equal to 26.
so the triangle is not a right triangle.

Question 19.

Answer:

Question 20.

Answer:
Yes, the triangle forms a right triangle.

Explanation:
In the above-given question,
given that,
the side lengths are 80 and 39.
the hypotenuse = 89.
89 x 89 = 80 x 80 + 39 x 39.
7921 = 6400 + 1521.
7921 = 7921.
7921 is equal to 7921.
so the triangle forms a right triangle.

In Exercises 21 – 28, verify that the segment lengths form a triangle. Is the triangle acute, right, or obtuse?

Question 21.
10, 11, and 14
Answer:

Question 22.
6, 8, and 10

Answer:
Yes, the triangle is forming a right triangle.

Explanation:
In the above-given question,
given that,
the side lengths are 8 and 6.
the hypotenuse = 10.
10 x 10 = 8 x 8 + 6 x 6.
100 = 64 + 36.
100 = 100.
100 is  equal to 100.
so the triangle is forming a right triangle.

Question 23.
12, 16, and 20
Answer:

Question 24.
15, 20, and 36

Answer:
Yes, the triangle is obtuse triangle.

Explanation:
In the above-given question,
given that,
the side lengths are 15 and 20.
the hypotenuse = 36.
36 x 36 = 20 x 20 + 15 x 15.
1296 = 400 + 225.
1296 > 625.
1296 is greater than 625.
so the triangle is not a obtuse triangle.

Question 25.
5.3, 6.7, and 7.8
Answer:

Question 26.
4.1, 8.2, and 12.2

Answer:
No, the triangle is obtuse triangle.

Explanation:
In the above-given question,
given that,
the side lengths are 4.1 and 8.2.
the hypotenuse = 12.2.
12.2 x 12.2 = 4.1 x 4.1 + 8.2 x 8.2`.
148.84 = 16.81 + 67.24.
148.84 > 84.05.
148.84 is greater than 84.05.
so the triangle is obtuse triangle.

Question 27.
24, 30, and 6√43
Answer:

Question 28.
10, 15 and 5√13

Answer:
Yes, the triangle is an acute triangle.

Explanation:
In the above-given question,
given that,
the side lengths are 10 and 5√13.
the hypotenuse = 15.
15 x 15 = 10 x 10 + 5√13 x 5√13.
225 = 100 + 34.81.
225 < 134.81.
225 is less than 134.81.
so the triangle is acute triangle.

Question 29.
MODELING WITH MATHEMATICS
In baseball, the lengths of the paths between consecutive bases are 90 feet, and the paths form right angles. The player on first base tries to steal second base. How far does the ball need to travel from home plate to second base to get the player out?
Answer:

Question 30.
REASONING
You are making a canvas frame for a painting using stretcher bars. The rectangular painting will be 10 inches long and 8 inches wide. Using a ruler, how can you be certain that the corners of the frame are 90°

Answer:
x = 12.8.

Explanation:
In the above-given question,
given that,
the side lengths are 10 and 8.
the hypotenuse = x.
X x X = 10 x 10 + 8 x 8.
X = 100 + 64.
X = 12.8.
In Exercises 31 – 34, find the area of the isosceles triangle.

Question 31.

Answer:

Question 32.

Answer:
The area of the Isosceles triangle = 12 ft.

Explanation:
In the above-given question,
given that,
base = 32 ft.
hypotenuse = 20ft.
a² + b² = c²
h x h + 16 x 16 = 20 x 20.
h x h + 256 = 400.
h x h = 400 – 256.
h x h = 144.
h = 12 ft.
so the area of the isosceles triangle = 12 ft.

Question 33.

Answer:

Question 34.

Answer:
The area of the Isosceles triangle = 48 m.

Explanation:
In the above-given question,
given that,
base = 28 m.
hypotenuse = 50 m.
a² + b² = c²
h x h + 14 x 14 = 50 x 50.
h x h + 196 = 2500.
h x h = 2500 – 196.
h x h = 2304.
h = 48 m.
so the area of the isosceles triangle = 48 m.

Question 35.
ANALYZING RELATIONSHIPS
Justify the Distance Formula using the Pythagorean Theorem (Thin. 9. 1).
Answer:

Question 36.
HOW DO YOU SEE IT?
How do you know ∠C is a right angle without using the Pythagorean Theorem (Theorem 9.1) ?

Answer:
Yes, the triangle is forming a right triangle.

Explanation:
In the above-given question,
given that,
the side lengths are 8 and 6.
the hypotenuse = 10.
10 x 10 = 8 x 8 + 6 x 6.
100 = 64 + 36.
100 = 100.
100 is  equal to 100.
so the triangle is forming a right triangle.

Question 37.
PROBLEM SOLVING
You are making a kite and need to figure out how much binding to buy. You need the binding for the perimeter of the kite. The binding Comes in packages of two yards. How many packages should you buy?

Answer:

Question 38.
PROVING A THEOREM
Use the Pythagorean Theorem (Theorem 9. 1) to prove the Hypotenuse-Leg (HL) Congruence Theorem (Theorem 5.9).

Answer:

Question 39.
PROVING A THEOREM
Prove the Converse of the Pythagorean Theorem (Theorem 9.2). (Hint: Draw ∆ABC with side lengths a, b, and c, where c is the length of the longest side. Then draw a right triangle with side lengths a, b, and x, where x is the length of the hypotenuse. Compare lengths c and x.)
Answer:

Question 40.
THOUGHT PROVOKING
Consider two integers m and n. where m > n. Do the following expressions produce a Pythagorean triple? If yes, prove your answer. If no, give a counterexample.
2mn, m² – n², m² + n²

Answer:

Question 41.
MAKING AN ARGUMENT
Your friend claims 72 and 75 Cannot be part of a pythagorean triple because 72² + 75² does not equal a positive integer squared. Is your friend correct? Explain your reasoning.
Answer:

Question 42.
PROVING A THEOREM
Copy and complete the proof of the pythagorean Inequalities Theorem (Theorem 9.3) when c² < a² + b².
Given In ∆ABC, c² < a² + b² where c is the length
of the longest side.
∆PQR has side lengths a, b, and x, where x is the length of the hypotenuse, and ∠R is a right angle.
Prove ∆ABC is an acute triangle.

Statements	Reasons
1. In ∆ABC, C2 < (12 + h2, where c is the length of the longest side. ∆PQR has side lengths a, b, and x, where x is the length of the hypotenuse, and ∠R is a right angle.	1. _____________________________
2. a2 + b2 = x2	2. _____________________________
3. c2 < r2	3. _____________________________
4. c < x	4. Take the positive square root of each side.
5. m ∠ R = 90°	5. _____________________________
6. m ∠ C < m ∠ R	6. Converse of the Hinge Theorem (Theorem 6.13)
7. m ∠ C < 90°	7. _____________________________
8. ∠C is an acute angle.	8. _____________________________
9. ∆ABC is an acute triangle.	9. _____________________________

Answer:

Question 43.
PROVING A THEOREM
Prove the Pythagorean Inequalities Theorem (Theorem 9.3) when c² > a² + b². (Hint: Look back at Exercise 42.)
Answer:

Maintaining Mathematical Proficiency

Simplify the expression by rationalizing the denominator.

Question 44.
\(\frac{7}{\sqrt{2}}\)

Answer:
7/√2 = 7√2 /2.

Explanation:
In the above-given question,
given that,
\(\frac{7}{\sqrt{2}}\) = 7/√2.
7/√2 = 7/√2  x √2 /√2 .
7 √2 /√4.
7√2 /2.

Question 45.
\(\frac{14}{\sqrt{3}}\)
Answer:

Question 46.
\(\frac{8}{\sqrt{2}}\)

Answer:
8/√2 = 8√2 /2.

Explanation:
In the above-given question,
given that,
\(\frac{8}{\sqrt{2}}\) = 8/√2.
8/√2 = 8/√2  x √2 /√2 .
8 √2 /√4.
8√2 /2.

Question 47.
\(\frac{12}{\sqrt{3}}\)
Answer:

9.2 Special Right Triangles

Exploration 1

Side Ratios of an Isosceles Right Triangle

Work with a partner:

a. Use dynamic geometry software to construct an isosceles right triangle with a leg length of 4 units.
Answer:

b. Find the acute angle measures. Explain why this triangle is called a 45° – 45° – 90° triangle.
Answer:

c. Find the exact ratios of the side lenghts (using square roots).
\(\frac{A B}{A C}\) = ____________
\(\frac{A B}{B C}\) = ____________
\(\frac{A B}{B C}\) = ____________
ATTENDING TO PRECISION
To be proficient in math, you need to express numerical answers with a degree of precision appropriate for the problem context.
Answer:

d. Repeat parts (a) and (c) for several other isosceles right triangles. Use your results to write a conjecture about the ratios of the side lengths of an isosceles right triangle.
Answer:

Exploration 2

Work with a partner.

a. Use dynamic geometry software to construct a right triangle with acute angle measures of 30° and 60° (a 30° – 60° – 90° triangle), where the shorter leg length is 3 units.

b. Find the exact ratios of the side lengths (using square roots).
\(\frac{A B}{A C}\) = ____________
\(\frac{A B}{B C}\) = ____________
\(\frac{A B}{B C}\) = ____________

Answer:

C. Repeat parts (a) and (b) for several other 30° – 60° – 90° triangles. Use your results to write a conjecture about the ratios of the side lengths of a 30° – 60° – 90° triangle.
Answer:

Communicate Your Answer

Question 3.
What is the relationship among the side lengths of 45°- 45° – 90° triangles? 30° – 60° – 90° triangles?
Answer:

Lesson 9.2 Special Right Triangles

Monitoring Progress

Find the value of the variable. Write your answer in simplest form.

Question 1.

Answer:
x = 4

Explanation:
(2√2)² = x² + x²
8 = 2x²
x² = 4
x = 4

Question 2.

Answer:
y = 2

Explanation:
y² = 2 + 2
y² = 4
y = 2

Question 3.

Answer:
x = 3, y = 2√3

Explanation:
longer leg = shorter leg • √3
x = √3 • √3
x = 3
hypotenuse = shorter leg • 2
= √3 • 2 = 2√3

Question 4

Answer:
h = 2√3

Explanation:
longer leg = shorter leg • √3
h = 2√3

Question 5.
The logo on a recycling bin resembles an equilateral triangle with side lengths of 6 centimeters. Approximate the area of the logo.

Answer:
Area is \(\frac { 1 }{ 3√3 } \)

Explanation:
Area = \(\frac { √3 }{ 4 } \) a²
= \(\frac { √3 }{ 4 } \)(6)²
= \(\frac { 1 }{ 3√3 } \)

Question 6.
The body of a dump truck is raised to empty a load of sand. How high is the 14-foot-long body from the frame when it is tipped upward by a 60° angle?

Answer:
28/3 ft high is the 14-foot-long body from the frame when it is tipped upward by a 60° angle.

Explanation:
Height of body at 90 degrees = 14 ft
Height of body at 1 degree = 14/90
Height of body at 60 degrees = 14 x 60/90
= 14 x 2/3
= 28/3 ft

Exercise 9.2 Special Right Triangles

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Name two special right triangles by their angle measures.
Answer:

Question 2.
WRITING
Explain why the acute angles in an isosceles right triangle always measure 45°.

Answer:
Because the acute angles of a right isosceles triangle must be congruent by the base angles theorem and complementary, their measures must be 90°/2 = 45°.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, find the value of x. Write your answer in simplest form.

Question 3.

Answer:

Question 4.

Answer:
x = 10

Explanation:
hypotenuse = leg • √2
x = 5√2 • √2
x = 10

Question 5.

Answer:

Question 6.

Answer:
x = \(\frac { 9 }{ √2 } \)

Explanation:
hypotenuse = leg • √2
9 = x • √2
x = \(\frac { 9 }{ √2 } \)

In Exercises 7 – 10, find the values of x and y. Write your answers in simplest form.

Question 7.

Answer:

Question 8.

Answer:
x = 3, y = 6

Explanation:
hypotenuse = 2 • shorter leg
y = 2 • 3
y = 6
longer leg = √3 • shorter leg
3√3 = √3x
x = 3

Question 9.

Answer:

Question 10.

Answer:
x = 18, y = 6√3

Explanation:
hypotenuse = 2 • shorter leg
12√3 = 2y
y = 6√3
longer leg = √3 • shorter leg
x = √3 . 6√3
x = 18

ERROR ANALYSIS
In Exercises 11 and 12, describe and correct the error in finding the length of the hypotenuse.

Question 11.

Answer:

Question 12.

Answer:
hypotenuse = leg • √2 = √5 . √2 = √10

In Exercises 13 and 14. sketch the figure that is described. Find the indicated length. Round decimal answers to the nearest tenth.

Question 13.
The side length of an equilateral triangle is 5 centimeters. Find the length of an altitude.
Answer:

Question 14.
The perimeter of a square is 36 inches. Find the length of a diagonal.

Answer:
The length of a diagonal is 9√2

Explanation:
Side of the square = 36/4 = 9
square diagonal = √2a = √2(9) = 9√2

In Exercises 15 and 16, find the area of the figure. Round decimal answers to the nearest tenth.

Question 15.

Answer:

Question 16.

Answer:
Area is 40√(1/3) sq m

Explanation:
longer leg = √3 • shorter leg
4 = √3 • shorter leg
shorter leg = 4/√3
h² = 16/3 + 16
h² = 16(4/3)
h = 8√(1/3)
Area of the parallelogram = 5(8√(1/3)) = 40√(1/3) sq m

Question 17.
PROBLEM SOLVING
Each half of the drawbridge is about 284 feet long. How high does the drawbridge rise when x is 30°? 45°? 60°?

Answer:

Question 18.
MODELING WITH MATHEMATICS
A nut is shaped like a regular hexagon with side lengths of 1 centimeter. Find the value of x. (Hint: A regular hexagon can be divided into six congruent triangles.)

Answer:
Side length = 1cm
A regular hexagon has six equal the side length. A line drawn from the centre to any vertex will have the same length as any side.
This implies the radius is equal to the side length.
As a result, when lines are drawn from the centre to each of the vertexes, a
regular hexagon is said to be made of six equilateral triangles.
From the diagram, x = 2× apothem
Apothem is the distance from the centre of a regular polygon to the midpoint of side.
Using Pythagoras theorem, we would get the apothem
Hypotenuse² = opposite² + adjacent²
1² = apothem² + (½)²
Apothem = √(1² -(½)²)
= √(1-¼) = √¾
Apothem = ½√3
x = 2× Apothem = 2 × ½√3
x = √3

Question 19.
PROVING A THEOREM
Write a paragraph proof of the 45°- 45°- 90° Triangle Theorem (Theorem 9.4).
Given ∆DEF is a 45° – 45° – 90° triangle.
Prove The hypotenuse is √2 times as long as each leg.

Answer:

Question 20.
HOW DO YOU SEE IT?
The diagram shows part of the wheel of Theodorus.

a. Which triangles, if any, are 45° – 45° – 90° triangles?

Answer:

b. Which triangles, if any, are 30° – 60° – 90° triangles?

Answer:

Question 21.
PROVING A THEOREM
Write a paragraph proof of the 30° – 60° – 90° Triangle Theorem (Theorem 9.5).
(Hint: Construct ∆JML congruent to ∆JKL.)
Given ∆JKL is a 30° 60° 9o° triangle.
Prove The hypotenuse is twice as long as the shorter leg, and the longer leg is √3 times as long as the shorter leg.

Answer:

Question 22.
THOUGHT PROVOKING
A special right triangle is a right triangle that has rational angle measures and each side length contains at most one square root. There are only three special right triangles. The diagram below is called the Ailles rectangle. Label the sides and angles in the diagram. Describe all three special right triangles.

Answer:

Question 23.
WRITING
Describe two ways to show that all isosceles right triangles are similar to each other.
Answer:

Question 24.
MAKING AN ARGUMENT
Each triangle in the diagram is a 45° – 45° – 90° triangle. At Stage 0, the legs of the triangle are each 1 unit long. Your brother claims the lengths of the legs of the triangles added are halved at each stage. So, the length of a leg of a triangle added in Stage 8 will be \(\frac{1}{256}\) unit. Is your brother correct? Explain your reasoning.

Answer:

Question 25.
USING STRUCTURE
ΔTUV is a 30° – 60° – 90° triangle. where two vertices are U(3, – 1) and V( – 3, – 1), \(\overline{U V}\) is the hypotenuse. and point T is in Quadrant I. Find the coordinates of T.
Answer:

Maintaining Mathematical Proficiency

Find the Value of x.

Question 26.
ΔDEF ~ ΔLMN

Answer:
x = 18

Explanation:
\(\frac { DE }{ LM } \) = \(\frac { DF }{ LN } \)
\(\frac { 12 }{ x } \) = \(\frac { 20 }{ 30 } \)
\(\frac { 12 }{ x } \) = \(\frac { 2 }{ 3 } \)
x = 12(\(\frac { 3 }{ 2 } \)) = 18

Question 27.
ΔABC ~ ΔQRS

Answer:

9.3 Similar Right Triangles

Exploration 1

Writing a Conjecture

a. Use dynamic geometry software to construct right ∆ABC, as shown. Draw \(\overline{C D}\) so that it is an altitude from the right angle to the hypotenuse of ∆ABC.

Answer:

b. The geometric mean of two positive numbers a and b is the positive number x that satisfies
\(\frac{a}{x}=\frac{x}{b}\)
x is the geometric mean of a and b.
Write a proportion involving the side lengths of ∆CBD and ∆ACD so that CD is the geometric mean of two of the other side lengths. Use similar triangles to justify your steps.
Answer:

c. Use the proportion you wrote in part (b) to find CD.
Answer:

d. Generalize the proportion you wrote in part (b). Then write a conjecture about how the geometric mean is related to the altitude from the right angle to the hypotenuse of a right triangle.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to understand and use stated assumptions, definitions, and previously established results in constructing arguments.
Answer:

Exploration 2

Comparing Geometric and Arithmetic Means

Work with a partner:
Use a spreadsheet to find the arithmetic mean and the geometric mean of several pairs of positive numbers. Compare the two means. What do you notice?

Answer:

Communicate Your Answer

Question 3.
How are altitudes and geometric means of right triangles related?
Answer:

Lesson 9.3 Similar Right Triangles

Monitoring progress

Identify the similar triangles.

Question 1.

Answer:
△QRS ~ △ QST

Question 2.

Answer:
△EFG ~ △ EHG

Question 3.

Answer:
△EGH ~ △EFG
\(\frac { EF }{ EG } \) = \(\frac { GF }{ GH } \)
\(\frac { 5 }{ 3 } \) = \(\frac { 4 }{ x } \)
x = 2.4

Question 4.

Answer:
△JLM ~ △LMK
\(\frac { JL }{ LM } \) = \(\frac { JM }{ KM } \)
\(\frac { 13 }{ 5 } \) = \(\frac { 12 }{ x } \)
x = 4.615

Find the geometric mean of the two numbers.

Question 5.
12 and 27

Answer:
x = √(12 x 27)
x = √324 = 18

Question 6.
18 and 54

Answer:
x = √(18 x 54) = √(972)
x = 31.17

Question 7.
16 and 18

Answer:
x = √(16 x 18) = √(288)
x = 16.970

Question 8.
Find the value of x in the triangle at the left.

Answer:
x = √(9 x 4)
x = 6

Question 9.
WHAT IF?
In Example 5, the vertical distance from the ground to your eye is 5.5 feet and the distance from you to the gym wall is 9 feet. Approximate the height of the gym wall.

Answer:
9² = 5.5 x w
81 = 5.5 x w
w = 14.72
The height of the wall = 14.72 + 5.5 = 20.22

Exercise 9.3 Similar Right Triangles

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
If the altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to the original triangle and ____________ .
Answer:

Question 2.
WRITING
In your own words, explain geometric mean.

Answer:
The geometric mean is the average value or mean that signifies the central tendency of set of numbers by finding the product of their values.

Monitoring progress and Modeling with Mathematics

In Exercises 3 and 4, identify the similar triangles.

Question 3.

Answer:

Question 4.

Answer:
△LKM ~ △LMN ~ △MKN

In Exercises 5 – 10, find the value of x.

Question 5.

Answer:

Question 6.

Answer:
x = 9.6

Explanation:
\(\frac { QR }{ SR } \) = \(\frac { SR }{ TS } \)
\(\frac { 20 }{ 16 } \) = \(\frac { 12 }{ x } \)
1.25 = \(\frac { 12 }{ x } \)
x = 9.6

Question 7.

Answer:

Question 8.

Answer:
x = 14.11

Explanation:
\(\frac { AB }{ AC } \) = \(\frac { BD }{ BC } \)
\(\frac { 16 }{ 34 } \) = \(\frac { x }{ 30 } \)
x = 14.11

Question 9.

Answer:

Question 10.

Answer:
x = 2.77

Explanation:
\(\frac { 5.8 }{ 3.5 } \) = \(\frac { 4.6 }{ x } \)
x = 2.77

In Exercises 11 – 18, find the geometric mean of the two numbers.

Question 11.
8 and 32
Answer:

Question 12.
9 and 16

Answer:
x = √(9 x 16)
x = 12

Question 13.
14 and 20
Answer:

Question 14.
25 and 35

Answer:
x = √(25 x 35)
x = 29.5

Question 15.
16 and 25
Answer:

Question 16.
8 and 28

Answer:
x = √(8 x 28)
x = 14.96

Question 17.
17 and 36
Answer:

Question 18.
24 and 45

Answer:
x = √(24 x 45)
x = 32.86

In Exercises 19 – 26. find the value of the variable.

Question 19.

Answer:

Question 20.

Answer:
y = √(5 x 8)
y = √40
y = 2√10

Question 21.

Answer:

Question 22.

Answer:
10 • 10 = 25 • x
100 = 25x
x = 4

Question 23.

Answer:

Question 24.

Answer:
b² = 16(16 + 6)
b² = 16(22) = 352
b = 18.76

Question 25.

Answer:

Question 26.

Answer:
x² = 8(8 + 2)
x² = 8(10) = 80
x = 8.9

ERROR ANALYSIS
In Exercises 27 and 28, describe and correct the error in writing an equation for the given diagram.

Question 27.

Answer:

Question 28.

Answer:
d² = g • e

MODELING WITH MATHEMATICS
In Exercises 29 and 30, use the diagram.

Question 29.
You want to determine the height of a monument at a local park. You use a cardboard square to line up the top and bottom of the monument, as shown at the above left. Your friend measures the vertical distance from the ground to your eye and the horizontal distance from you to the monument. Approximate the height of the monument.
Answer:

Question 30.
Your classmate is standing on the other side of the monument. She has a piece of rope staked at the base of the monument. She extends the rope to the cardboard square she is holding lined up to the top and bottom of the monument. Use the information in the diagram above to approximate the height of the monument. Do you get the same answer as in Exercise 29? Explain your reasoning.

Answer:

MATHEMATICAL CONNECTIONS
In Exercises 31 – 34. find the value(s) of the variable(s).

Question 31.

Answer:

Question 32.

Answer:
\(\frac { 6 }{ b + 3 } \) = \(\frac { 8 }{ 6 } \)
36 = 8(b + 3)
36 = 8b + 24
8b = 12
b = \(\frac { 3 }{ 2 } \)

Question 33.

Answer:

Question 34.

Answer:
x = 42.66, y = 40, z = 53

Explanation:
\(\frac { 24 }{ 32 } \) = \(\frac { 32 }{ x } \)
0.75 = \(\frac { 32 }{ x } \)
x = 42.66
y = √24² + 32²
y = √576 + 1024 = 40
z = √42.66² + 32² = √1819.87 + 1024 = 53

Question 35.
REASONING
Use the diagram. Decide which proportions are true. Select all that apply.

(A) \(\frac{D B}{D C}=\frac{D A}{D B}\)
(B) \(\frac{B A}{C B}=\frac{C B}{B D}\)
(C) \(\frac{C A}{B \Lambda}=\frac{B A}{C A}\)
(D) \(\frac{D B}{B C}=\frac{D A}{B A}\)
Answer:

Question 36.
ANALYZING RELATIONSHIPS
You are designing a diamond-shaped kite. You know that AD = 44.8 centimeters, DC = 72 centimeters, and AC = 84.8 centimeters. You Want to use a straight crossbar \(\overline{B D}\). About how long should it be? Explain your reasoning.

Answer:
BD = 76.12

Explanation:
AD = 44.8 cm, DC = 72 cm, and AC = 84.8 cm
Two disjoint pairs of consecutive sides are congruent.
So, AD = AB = 44.8 cm
DC = BC = 72 cm
The diagonals are perpendicular.
So, AC ⊥ BD
AC = AO + OC
AX = x + y = 84.8 — (i)
Perpendicular bisects the diagonal BD into equal parts let it be z.
BD = BO + OD
BD = z + z
Using pythagorean theorem
44.8² = x² + z² —- (ii)
72² = y² + z² —– (iii)
Subtract (ii) and (iii)
72² – 44.8² = y²+ z² – x² – z²
5184 – 2007.04 = (x + y) (x – y)
3176.96 = (84.8)(x – y)
37.464 = x – y —- (iv)
Add (i) & (iv)
x + y + x – y = 84.8 + 37.464
2x = 122.264
x = 61.132
x + y = 84.8
61.132 + y = 84.8
y = 23.668
44.8² = x² + z²
z = 38.06
BD = z + z
BD = 76.12

Question 37.
ANALYZING RELATIONSHIPS
Use the Geometric Mean Theorems (Theorems 9.7 and 9.8) to find AC and BD.

Answer:

Question 38.
HOW DO YOU SEE IT?
In which of the following triangles does the Geometric Mean (Altitude) Theorem (Theorem 9.7) apply?
(A)

(B)

(C)

(D)

Answer:

Question 39.
PROVING A THEOREM
Use the diagram of ∆ABC. Copy and complete the proof of the Pythagorean Theorem (Theorem 9. 1).

Given In ∆ABC, ∆BCA is a right angle.
Prove c² = a² + b²

Statements	Reasons
1. In ∆ABC, ∠BCA is a right angle.	1. ________________________________
2. Draw a perpendicular segment (altitude) from C to \(\overline{A B}\).	2. Perpendicular Postulate (Postulate 3.2)
3. ce = a2 and cf = b2	3. ________________________________
4. ce + b2 = ___  + b2	4. Addition Property of Equality
5. ce + cf = a2 + b2	5. ________________________________
6. c(e + f) a2 + b2	6. ________________________________
7. e + f = ________	7. Segment Addition Postulate (Postulate 1.2)
8. c  • c = a2 + b2	8. ________________________________
9. c2 = a2 + b2	9. Simplify.

Answer:

Question 40.
MAKING AN ARGUMENT
Your friend claims the geometric mean of 4 and 9 is 6. and then labels the triangle, as shown. Is your friend correct? Explain your reasoning.

Answer:
G.M = √(4 x 9) = √36 = 6
My friend is correct.

In Exercises 41 and 42, use the given statements to prove the theorem.

Gien ∆ABC is a right triangle.
Altitude \(\overline{C D}\) is dravn to hypotenuse \(\overline{A B}\).

Question 41.
PROVING A THEOREM
Prove the Geometric Mean (Altitude) Theorem (Theorem 9.7) b showing that CD² = AD • BD.
Answer:

Question 42.
PROVING A THEOREM
Prove the Geometric Mean ( Leg) Theorem (Theorem 9.8) b showing that CB² = DB • AB and AC² = AD • AB.

Answer:

Question 43.
CRITICAL THINKING
Draw a right isosceles triangle and label the two leg lengths x. Then draw the altitude to the hypotenuse and label its length y. Now, use the Right Triangle Similarity Theorem (Theorem 9.6) to draw the three similar triangles from the image and label an side length that is equal to either x or y. What can you conclude about the relationship between the two smaller triangles? Explain your reasoning.
Answer:

Question 44.
THOUGHT PROVOKING
The arithmetic mean and geometric mean of two nonnegative numbers x and y are shown.
arithmetic mean = \(\frac{x+y}{2}\)
geometric mean = \(\sqrt{x y}\)
Write an inequality that relates these two means. Justify your answer.

Answer:

Question 45.
PROVING A THEOREM
Prove the Right Triangle Similarity Theorem (Theorem 9.6) by proving three similarity statements.
Given ∆ABC is a right triangle.
Altitude \(\overline{C D}\) is drawn to hvpotenuse \(\overline{A B}\).
Prove ∆CBD ~ ∆ABC, ∆ACD ~ ∆ABC,
∆CBD ~ ∆ACD
Answer:

Maintaining Mathematical proficiency

Solve the equation for x.

Question 46.
13 = \(\frac{x}{5}\)

Answer:
13 = \(\frac{x}{5}\)
x = 65

Question 47.
29 = \(\frac{x}{4}\)
Answer:

Question 48.
9 = \(\frac{78}{x}\)

Answer:
9 = \(\frac{78}{x}\)
9x = 78
x = 8.6

Question 49.
30 = \(\frac{115}{x}\)
Answer:

9.1 to 9.3 Quiz

Find the value of x. Tell whether the side lengths form a Pythagorean triple.

Question 1.

Answer:
x = 15

Explanation:
x² = 9² + 12²
x² = 81 + 144
x² = 225
x = 15

Question 2.

Answer:
x = 10.63

Explanation:
x² = 7² + 8² = 49 + 64
x = √113
x = 10.63

Question 3.

Answer:
x = 4√3

Explanation:
8² = x² + 4²
64 = x² + 16
x² = 48
x = 4√3

Verify that the segment lengths form a triangle. Is the triangle acute, right, or obtuse?
(Section 9.1)
Question 4.
24, 32, and 40

Answer:
Triangle is a right angle trinagle.

Explanation:
40² = 1600
24² + 32² = 576 + 1024 = 1600
40² = 24² + 32²
So, the triangle is a right angle trinagle.

Question 5.
7, 9, and 13

Answer:
Triangle is an obtuse trinagle.

Explanation:
13² = 169
7² + 9² = 49 + 81 = 130
13² > 7² + 9²
So, the triangle is an obtuse trinagle.

Question 6.
12, 15, and 10√3

Answer:
Triangle is an acute trinagle.

Explanation:
15² = 225
12² + (10√3)² = 144 + 300 = 444
15² < 12² + (10√3)²
So, the triangle is an acute trinagle.

Find the values of x and y. Write your answers in the simplest form.

Question 7.

Answer:
x = 6, y = 6√2

Explanation:
x = 6
hypotenuse = leg • √2
y = 6√2

Question 8.

Answer:
y = 8√3, x = 16

Explanation:
longer leg = shorter leg • √3
y = 8√3
x² = 8² + (8√3)² = 64 + 192
x = 16

Question 9.

Answer:
x = 5√2, y = 5√6

Explanation:
longer leg = shorter leg • √3
y = x√3
y = 5√6
hypotenuse = shorter leg • 2
10√2 = 2x
x = 5√2

Find the geometric mean of the two numbers.
Question 10.
6 and 12

Answer:
G.M = √(6 • 12) = 6√2

Question 11.
15 and 20

Answer:
G.M = √(15 • 20) = 10√3

Question 12.
18 and 26

Answer:
G.M = √(18 • 26) = 6√13

Identify the similar right triangles. Then find the value of the variable.

Question 13.

Answer:
x = √(8 • 4)
x = 4√2

Question 14.

Answer:
y = √(9 • 6) = 3√6

Question 15.

Answer:

Question 16.
Television sizes are measured by the length of their diagonal. You want to purchase a television that is at least 40 inches. Should you purchase the television shown? Explain your reasoning.

Answer:
x² = 20.25² + 36²
x² = 410.0625 + 1296 = 1706.0625
x = 41.30
Yes, i will purchase the television.

Question 17.
Each triangle shown below is a right triangle.

a. Are any of the triangles special right triangles? Explain your reasoning.
Answer:
A is a similar triangle.

b. List all similar triangles. if any.
Answer:
B, C and D, E are similar triangles.

c. Find the lengths of the altitudes of triangles B and C.
Answer:
B altitude = √(9 + 27) = 6
C altitude = √(36 + 72) = 6√3

9.4 The Tangent Ratio

Exploration 1

Calculating a Tangent Ratio

Work with a partner

a. Construct ∆ABC, as shown. Construct segments perpendicular to \(\overline{A C}\) to form right triangles that share vertex A and arc similar to ∆ABC with vertices, as shown.

Answer:

b. Calculate each given ratio to complete the table for the decimal value of tan A for each right triangle. What can you Conclude?

Answer:

Exploration 2

Using a calculator

Work with a partner: Use a calculator that has a tangent key to calculate the tangent of 36.87°. Do you get the same result as in Exploration 1? Explain.
ATTENDING TO PRECISION
To be proficient in math, you need to express numerical answers with a degree of precision appropriate for the problem context.
Answer:

Communicate Your Answer

Question 3.
Repeat Exploration 1 for ∆ABC with vertices A(0, 0), B(8, 5), and C(8, 0). Construct the seven perpendicular segments so that not all of them intersect \(\overline{A C}\) at integer values of x. Discuss your results.
Answer:

Question 4.
How is a right triangle used to find the tangent of an acute angle? Is there a unique right triangle that must be used?
Answer:

Lesson 9.4 The Tangent Ratio

Monitoring progress

Find tan J and tan K. Write each answer as a fraction and as a decimal rounded to four places.

Question 1.

Answer:
tan K = \(\frac { opposite side }{ adjacent side } \) = \(\frac { JL }{ KL } \)
= \(\frac { 32 }{ 24 } \) = \(\frac { 4 }{ 3 } \) = 1.33
tan J = \(\frac { KL }{ JL } \) = \(\frac { 24 }{ 32 } \) = \(\frac { 3 }{ 4 } \) = 0.75

Question 2.

Answer:
tan K = \(\frac { LJ }{ LK } \) = \(\frac { 15 }{ 8 } \)
tan J = \(\frac { LK }{ LJ } \) = \(\frac { 8 }{ 15 } \)

Find the value of x. Round your answer to the nearest tenth.

Question 3.

Answer:
Tan 61 = \(\frac { 22 }{ x } \)
1.804 = \(\frac { 22 }{ x } \)
x = 12.1951

Question 4.

Answer:
tan 56 = \(\frac { x }{ 13 } \)
1.482 = \(\frac { x }{ 13 } \)
x = 19.266

Question 5.
WHAT IF?
In Example 3, the length of the shorter leg is 5 instead of 1. Show that the tangent of 60° is still equal to √3.

Answer:
longer leg = shorter leg • √3
= 5√3
tan 60 = \(\frac { 5√3 }{ 5 } \)
= √3

Question 6.
You are measuring the height of a lamppost. You stand 40 inches from the base of the lamppost. You measure the angle ot elevation from the ground to the top of the lamppost to be 70°. Find the height h of the lamppost to the nearest inch.

Answer:
tan 70 = \(\frac { h }{ 40 } \)
2.7474 = \(\frac { h }{ 40 } \)
h = 109.896 in

Exercise 9.4 The Tangent Ratio

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
The tangent ratio compares the length of _________ to the length of ___________ .
Answer:

Question 2.
WRITING
Explain how you know the tangent ratio is constant for a given angle measure.

Answer:
When two triangles are similar, the corresponding sides are proportional which makes the ratio constant for a given acute angle measurement.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, find the tangents of the acute angles in the right triangle. Write each answer as a fraction and as a decimal rounded to four decimal places.

Question 3.

Answer:

Question 4.

Answer:
tan F = \(\frac { DE }{ EF } \) = \(\frac { 24 }{ 7 } \)
tan D = \(\frac { EF }{ DE } \) = \(\frac { 7 }{ 24 } \)

Question 5.

Answer:

Question 6.

Answer:
tan K = \(\frac { JL }{ LK } \) = \(\frac { 3 }{ 5 } \)
tan J = \(\frac { LK }{ JL } \) = \(\frac { 5 }{ 3 } \)

In Exercises 7 – 10, find the value of x. Round your answer to the nearest tenth.

Question 7.

Answer:

Question 8.

Answer:
tan 27 = \(\frac { x }{ 15 } \)
0.509 = \(\frac { x }{ 15 } \)
x = 7.635

Question 9.

Answer:

Question 10.

Answer:
tan 37 = \(\frac { 6 }{ x } \)
0.753 = \(\frac { 6 }{ x } \)
x = 7.968

ERROR ANALYSIS
In Exercises 11 and 12, describe the error in the statement of the tangent ratio. Correct the error if possible. Otherwise, write not possible.

Question 11.

Answer:

Question 12.

Answer:

In Exercises 13 and 14, use a special right triangle to find the tangent of the given angle measure.

Question 13.
45°
Answer:

Question 14.
30°

Answer:
tan 30° = \(\frac { 1 }{ √3 } \)

Question 15.
MODELING WITH MATHEMATICS
A surveyor is standing 118 Feet from the base of the Washington Monument. The surveyor measures the angle of elevation from the ground to the top of the monument to be 78°. Find the height h of the Washington Monument to the nearest foot.

Answer:

Question 16.
MODELING WITH MATHEMATICS
Scientists can measure the depths of craters on the moon h looking at photos of shadows. The length of the shadow cast by the edge of a crater is 500 meters. The angle of elevation of the rays of the Sun is 55°. Estimate the depth d of the crater.

Answer:
tan 55 = \(\frac { d }{ 500 } \)
1.428 = \(\frac { d }{ 500 } \)
d = 714 m
The depth of the crater is 714 m

Question 17.
USING STRUCTURE
Find the tangent of the smaller acute angle in a right triangle with side lengths 5, 12, and 13.
Answer:

Question 18.
USING STRUCTURE
Find the tangent 0f the larger acute angle in a right triangle with side lengths 3, 4, and 5.

Answer:
tan x = \(\frac { 4 }{ 3 } \)

Question 19.
REASONING
How does the tangent of an acute angle in a right triangle change as the angle measure increases? Justify your answer.
Answer:

Question 20.
CRITICAL THINKING
For what angle measure(s) is the tangent of an acute angle in a right triangle equal to 1? greater than 1? less than 1? Justify your answer.

Answer:
In order for the tangent of an angle to equal 1, the opposite and adjacent sides of a right triangle must be the same. This means the right triangle is an isosceles right triangle so the angles are 45 – 45 – 90. The acute angle must be 1. In order for the tangent to be greater than 1, the opposite side must be greater than the adjacent side. This means the angle must be between 45 and 90 degrees. If the tangent is less than 1, this means the opposite side must be smaller than the adjacent side. The acute angle must be between 0 and 45.

Question 21.
MAKING AN ARGUMENT
Your family room has a sliding-glass door. You want to buy an awning for the door that will be just long enough to keep the Sun out when it is at its highest point in the sky. The angle of elevation of the rays of the Sun at this points is 70°, and the height of the door is 8 feet. Your sister claims you can determine how far the overhang should extend by multiplying 8 by tan 70°. Is your sister correct? Explain.

Answer:

Question 22.
HOW DO YOU SEE IT?
Write expressions for the tangent of each acute angle in the right triangle. Explain how the tangent of one acute angle is related to the tangent of the other acute angle. What kind of angle pair is ∠A and ∠B?

Answer:
tan A = \(\frac { BC }{ AC } \) = \(\frac { a }{ b } \)
tan B = \(\frac { AC }{ BC } \) = \(\frac { b }{ a } \)

Question 23.
REASONING
Explain why it is not possible to find the tangent of a right angle or an obtuse angle.
Answer:

Question 24.
THOUGHT PROVOKING
To create the diagram below. you begin with an isosceles right triangle with legs 1 unit long. Then the hypotenuse of the first triangle becomes the leg of a second triangle, whose remaining leg is 1 unit long. Continue the diagram Until you have constructed an angle whose tangent is \(\frac{1}{\sqrt{6}}\). Approximate the measure of this angle.

Answer:

Question 25.
PROBLEM SOLVING
Your class is having a class picture taken on the lawn. The photographer is positioned 14 feet away from the center of the class. The photographer turns 50° to look at either end of the class.

a. What is the distance between the ends of the class?
b. The photographer turns another 10° either way to see the end of the camera range. If each student needs 2 feet of space. about how many more students can fit at the end of each row? Explain.
Answer:

Question 26.
PROBLEM SOLVING
Find the perimeter of the figure. where AC = 26, AD = BF, and D is the midpoint of \(\overline{A C}\).

Answer:

Maintaining Mathematical proficiency

Find the value of x.

Question 27.

Answer:

Question 28.

Answer:
longer side = shorter side • √3
7 = x√3
x = \(\frac { 7 }{ √3 } \)
x = 4.04

Question 29.

Answer:

9.5 The Sine and Cosine Ratios

Exploration 1

Work with a partner: Use dynamic geometry software.

a. Construct ∆ABC, as shown. Construct segments perpendicular to \(\overline{A C}\) to form right triangles that share vertex A arid are similar to ∆ABC with vertices, as shown.

Answer:

b. Calculate each given ratio to complete the table for the decimal values of sin A and cos A for each right triangle. What can you conclude?

Answer:

Communicate Your Answer

Question 2.
How is a right triangle used to find the sine and cosine of an acute angle? Is there a unique right triangle that must be used?
Answer:

Question 3.
In Exploration 1, what is the relationship between ∠A and ∠B in terms of their measures’? Find sin B and cos B. How are these two values related to sin A and cos A? Explain why these relationships exist.
LOOKING FOR STRUCTURE
To be proficient in math, you need to look closely to discern a pattern or structure.
Answer:

Lesson 9.5 The Sine and Cosine Ratios

Monitoring Progress

Question 1.
Find sin D, sin F, cos D, and cos F. Write each answer as a fraction and as a decimal rounded to four places.

Answer:
sin D = \(\frac { 7 }{ 25 } \)
sin F = \(\frac { 24 }{ 25 } \)
cos D = \(\frac { 24 }{ 25 } \)
cos F = \(\frac { 7 }{ 25 } \)

Explanation:
sin D = \(\frac { EF }{ DF } \) = \(\frac { 7 }{ 25 } \)
sin F = \(\frac { DE }{ DF } \) = \(\frac { 24 }{ 25 } \)
cos D = \(\frac { DE }{ DF } \) = \(\frac { 24 }{ 25 } \)
cos F = \(\frac { EF }{ DF } \) = \(\frac { 7 }{ 25 } \)

Question 2.
Write cos 23° in terms of sine.

Answer:
cos X = sin(90 – X)
cos 23° = sin (90 – 23)
= sin(67)
So, cos 23° = sin 67°

Question 3.
Find the values of u and t using sine and cosine. Round your answers to the nearest tenth.

Answer:
t = 7.2, u = 3.3

Explanation:
sin 65 = \(\frac { t }{ 8 } \)
0.906 = \(\frac { t }{ 8 } \)
t = 7.2
cos 65 = \(\frac { u }{ 8 } \)
0.422 x 8 = u
u = 3.3

Question 4.
Find the sine and cosine of a 60° angle.

Answer:

sin 60° = \(\frac { √3 }{ 2 } \)
cos 60° = \(\frac { 1 }{ 2 } \)

Question 5.
WHAT IF?
In Example 6, the angle of depression is 28°. Find the distance x you ski down the mountain to the nearest foot.

Answer:
sin 28° = \(\frac { 1200 }{ x } \)
x = \(\frac { 1200 }{ 0.469 } \)
x = 2558.6

Exercise 9.5 The Sine and Cosine Ratios

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
The sine raio compares the length of ______________ to the length of _____________
Answer:

Question 2.
WHICH ONE DOESN’T BELONG?
Which ratio does not belong with the other three? Explain your reasoning.

sin B

Answer:
sin B = \(\frac { AC }{ BC } \)

cos C

Answer:
cos C = \(\frac { AC }{ BC } \)

tan B

Answer:
tan B = \(\frac { AC }{ AB } \)

\(\frac{A C}{B C}\)

Answer:
\(\frac{A C}{B C}\) = sin B

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 8, find sin D, sin E, cos D, and cos E. Write each answer as a Fraction and as a decimal rounded to four places.

Question 3.

Answer:

Question 4.

Answer:
sin D = \(\frac { 35 }{ 37 } \)
sin E = \(\frac { 12 }{ 37 } \)
cos D = \(\frac { 12 }{ 37 } \)
cos E = \(\frac { 35 }{ 37 } \)

Question 5.

Answer:

Question 6.

Answer:
sin D = \(\frac { 36 }{ 45 } \)
sin E = \(\frac { 27 }{ 45 } \)
cos D = \(\frac { 27 }{ 45 } \)
cos E = \(\frac { 36 }{ 45 } \)

Question 7.

Answer:

Question 8.

Answer:
sin D = \(\frac { 8 }{ 17 } \)
sin E = \(\frac { 15 }{ 17 } \)
cos D = \(\frac { 15 }{ 17 } \)
cos E = \(\frac { 8 }{ 17 } \)

In Exercises 9 – 12. write the expression in terms of cosine.

Question 9.
sin 37°
Answer:

Question 10.
sin 81°

Answer:
sin 81° = cos(90° – 81°) = cos9°

Question 11.
sin 29°
Answer:

Question 12.
sin 64°

Answer:
sin 64° = cos(90° – 64°) = cos 26°

In Exercise 13 – 16, write the expression in terms of sine.

Question 13.
cos 59°
Answer:

Question 14.
cos 42°

Answer:
cos 42° = sin(90° – 42°) = sin 48°

Question 15.
cos 73°
Answer:

Question 16.
cos 18°

Answer:
cos 18° = sin(90° – 18°) = sin 72°

In Exercises 17 – 22, find the value of each variable using sine and cosine. Round your answers to the nearest tenth.

Question 17.

Answer:

Question 18.

Answer:
p = 30.5, q = 14.8

Explanation:
sin 64° = \(\frac { p }{34 } \)
p = 0.898 x 34
p = 30.5
cos 64° = \(\frac { q }{ 34 } \)
q = 0.4383 x 34
q = 14.8

Question 19.

Answer:

Question 20.

Answer:
s = 17.7, r = 19

Explanation:
sin 43° = \(\frac { s }{26 } \)
s = 0.681 x 26
s = 17.7
cos 43° = \(\frac { r }{ 26 } \)
r = 0.731 x 26
r = 19

Question 21.

Answer:

Question 22.

Answer:
m = 6.7, n = 10.44

Explanation:
sin 50° = \(\frac { 8 }{n } \)
0.766 = \(\frac { 8 }{n } \)
n = 10.44
cos 50° = \(\frac { m }{ n } \)
0.642 = \(\frac { m }{ 10.44 } \)
m = 6.7

Question 23.
REASONING
Which ratios are equal? Select all that apply.

sin X

cos X

sin Z

cos Z
Answer:

Question 24.
REASONING
Which ratios arc equal to \(\frac{1}{2}\) Select all

sin L

Answer:
sin L = \(\frac { 2 }{ 4 } \) = \(\frac { 1 }{ 2 } \)

cos L

Answer:
cos L = \(\frac { 2√3 }{ 4 } \) = \(\frac { √3 }{ 2 } \)

sin J

Answer:
sin J = \(\frac { 2√3 }{ 4 } \) = \(\frac { √3 }{ 2 } \)

cos J

Answer:
cos J = \(\frac { 2 }{ 4 } \) = \(\frac { 1 }{ 2 } \)

Question 25.
ERROR ANALYSIS
Describe and correct the error in finding sin A.

Answer:

Question 26.
WRITING
Explain how to tell which side of a right triangle is adjacent to an angle and which side is the hypotenuse.
Answer:

Question 27.
MODELING WITH MATHEMATICS
The top of the slide is 12 feet from the ground and has an angle of depression of 53°. What is the length of the slide?

Answer:

Question 28.
MODELING WITH MATHEMATICS
Find the horizontal distance x the escalator covers.

Answer:
cos 41 = \(\frac { x }{ 26 } \)
0.754 = \(\frac { x }{ 26 } \)
x = 19.6 ft

Question 29.
PROBLEM SOLVING
You are flying a kite with 20 feet of string extended. The angle of elevation from the spool of string to the kite is 67°.
a. Draw and label a diagram that represents the situation.
b. How far off the ground is the kite if you hold the spool 5 feet off the ground? Describe how the height where you hold the spool affects the height of the kite.
Answer:

Question 30.
MODELING WITH MATHEMATICS
Planes that fly at high speeds and low elevations have radar s sterns that can determine the range of an obstacle and the angle of elevation to the top of the obstacle. The radar of a plane flying at an altitude of 20,000 feet detects a tower that is 25,000 feet away. with an angle of elevation of 1°

a. How many feet must the plane rise to pass over the tower?

Answer:
sin 1 = \(\frac { h }{ 25000 } \)
0.017 = \(\frac { h }{ 25000 } \)
h = 425 ft
425 ft the plane rise to pass over the tower

b. PIanes Caillot come closer than 1000 feet vertically to any object. At what altitude must the plane fly in order to pass over the tower?

Answer:

Question 31.
MAKING AN ARGUMENT
Your friend uses che equation sin 49° = \(\frac{x}{16}\) to find BC. Your cousin uses the equation cos 41° = \(\frac{x}{16}\) to find BC. Who is correct? Explain your reasoning.

Answer:

Question 32.
WRITING
Describe what you must know about a triangle in order to use the sine ratio and what you must know about a triangle in order to use the cosine ratio

Answer:
sin = \(\frac { opposite side }{ hypotenuse } \)
cos = \(\frac { adjacent side }{ hypotenuse } \)

Question 33.
MATHEMATICAL CONNECTIONS
If ∆EQU is equilateral and ∆RGT is a right triangle with RG = 2, RT = 1. and m ∠ T = 90°, show that sin E = cos G.
Answer:

Question 34.
MODELING WITH MATHEMATICS
Submarines use sonar systems, which are similar to radar systems, to detect obstacles, Sonar systems use sound to detect objects under water.

a. You are traveling underwater in a submarine. The sonar system detects an iceberg 4000 meters a head, with an angle of depression of 34° to the bottom of the iceberg. How many meters must the submarine lower to pass under the iceberg?

Answer:
tan 34 = \(\frac { x }{ 4000 } \)
.674 = \(\frac { x }{ 4000 } \)
x = 2696

b. The sonar system then detects a sunken ship 1500 meters ahead. with an angle of elevation of 19° to the highest part of the sunken ship. How many meters must the submarine rise to pass over the sunken ship?

Answer:
tan 19 = \(\frac { x }{ 1500 } \)
0.344 = \(\frac { x }{ 1500 } \)
x = 516 m

Question 35.
ABSTRACT REASONING
Make a conjecture about how you could use trigonometric ratios to find angle measures in a triangle.
Answer:

Question 36.
HOW DO YOU SEE IT?
Using only the given information, would you use a sine ratio or a cosine ratio to find the length of the hypotenuse? Explain your reasoning.

Answer:
sin 29 = \(\frac { 9 }{ x } \)
0.48 = \(\frac { 9 }{ x } \)
x = 18.75
The length of hypotenuse is 18.75

Question 37.
MULTIPLE REPRESENTATIONS
You are standing on a cliff above an ocean. You see a sailboat from your vantage point 30 feet above the ocean.

a. Draw and label a diagram of the situation.
b. Make a table showing the angle of depression and the length of your line of sight. Use the angles 40°, 50°, 60°, 70°, and 80°.
c. Graph the values you found in part (b), with the angle measures on the x-axis.
d. Predict the length of the line of sight when the angle of depression is 30°.
Answer:

Question 38.
THOUGHT PROVOKING
One of the following infinite series represents sin x and the other one represents cos x (where x is measured in radians). Which is which? Justify your answer. Then use each series to approximate the sine and cosine of \(\frac{\pi}{6}\).
(Hints: π = 180°; 5! = 5 • 4 • 3 • 2 • 1; Find the values that the sine and cosine ratios approach as the angle measure approaches zero).
a.

Answer:
For x = 0
0 – \(\frac { 0³ }{ 3! } \) + \(\frac { 0⁵ }{ 5! } \) – \(\frac { 0⁷ }{ 7! } \) + . . . = 0
sin x = x – \(\frac { x³ }{ 3! } \) + \(\frac { x⁵ }{ 5! } \) – \(\frac { x⁷ }{ 7! } \) + . . .
sin \(\frac { π }{ 6 } \) = 0.5

b.

Answer:
1 – \(\frac { 1² }{ 2! } \) + \(\frac { 1⁴ }{ 4! } \) – \(\frac { 1⁶ }{ 6! } \) + . .  = 1
cos x =x1 – \(\frac { x² }{ 2! } \) + \(\frac { x⁴ }{ 4! } \) – \(\frac { x⁶ }{ 6! } \) + . .
cos \(\frac { π }{ 6 } \) = 0.86

Question 39.
CRITICAL THINKING
Let A be any acute angle of a right triangle. Show that
(a) tan A = \(\frac{\sin A}{\cos A}\) and
(b) (sin A)² + (cos A)² = 1.
Answer:

Question 40.
CRITICAL THINKING
Explain why the area ∆ ABC in the diagram can be found using the formula Area = \(\frac{1}{2}\) ab sin C. Then calculate the area when a = 4, b = 7, and m∠C = 40°:

Answer:
Area = \(\frac{1}{2}\) ab sin C
= \(\frac{1}{2}\) (4 x 7) sin 40°
= 14 x 0.642
= 8.988

Maintaining Mathematical Proficiency

Find the value of x. Tell whether the side lengths form a Pythagorean triple.

Question 41.

Answer:

Question 42.

Answer:
x = 12√2

Explanation:
c² = a² + b²
x² = 12² + 12²
x² = 144 + 144
x² = 288
x = 12√2

Question 43.

Answer:

Question 44.

Answer:
x = 6√2

Explanation:
c² = a² + b²
9² = x² + 3²
81 = x² + 9
x² = 81 – 9
x = 6√2

9.6 Solving Right Triangles

Exploration 1

Solving Special Right Triangles

Work with a partner. Use the figures to find the values of the sine and cosine of ∠A and ∠B. Use these values to find the measures of ∠A and ∠B. Use dynamic geometry software to verify your answers.
a.

Answer:

b.

Answer:

Exploration 2

Solving Right Triangles

Work with a partner: You can use a calculator to find the measure of an angle when you know the value of the sine, cosine, or tangent of the rule. Use the inverse sine, inverse cosine, 0r inverse tangent feature of your calculator to approximate the measures of ∠A and ∠B to the nearest tenth of a degree. Then use dynamic geometry software to verify your answers.
ATTENDING TO PRECISION
To be proficient in math, you need to calculate accurately and efficiently, expressing numerical answers with a degree of precision appropriate for the problem context.
a.

Answer:

b.

Answer:

Communicate Your Answer

Question 3.
When you know the lengths of the sides of a right triangle, how can you find the measures of the two acute angles?
Answer:

Question 4.
A ladder leaning against a building forms a right triangle with the building and the ground. The legs of the right triangle (in meters) form a 5-12-13 Pythagorean triple. Find the measures of the two acute angles to the nearest tenth of a degree.
Answer:

Lesson 9.6 Solving Right Triangles

Determine which of the two acute angles has the given trigonometric ratio.

Question 1.
The sine of the angle is \(\frac{12}{13}\).

Answer:
Sin E = \(\frac{12}{13}\)
m∠E = sin^-1(\(\frac{12}{13}\)) = 67.3°

Question 2.
The tangent of the angle is \(\frac{5}{12}\)

Answer:
tan F = \(\frac{5}{12}\)
m∠F = tan^-1(\(\frac{5}{12}\)) = 22.6°

Let ∠G, ∠H, and ∠K be acute angles. Use a calculator to approximate the measures of ∠G, ∠H, and ∠K to the nearest tenth of a degree.

Question 3.
tan G = 0.43

Answer:
∠G = inverse tan of 0.43 = 23.3°

Question 4.
sin H = 0.68

Answer:
∠H = inverse sin of 0.68 = 42.8°

Question 5.
cos K = 0.94

Answer:
∠K = inverse cos of 0.94 = 19.9°

Solve the right triangle. Round decimal answers to the nearest tenth.

Question 6.

Answer:
DE = 29, ∠D = 46.05°, ∠E = 42.84°

Explanation:
c² = a² + b²
x² = 20² + 21²
x² = 400 + 441
x² = 841
x = 29
sin D = \(\frac { 21 }{ 29 } \)
∠D = 46.05
sin E = \(\frac { 20 }{ 29 } \)
∠E = 42.84

Question 7.

Answer:
GJ = 60, ∠G = 56.09°, ∠H = 33.3°

Explanation:
c² = a² + b²
109² = 91² + x²
x² = 11881 – 8281
x² = 3600
x = 60
sin G = \(\frac { 91 }{ 109 } \)
∠G = 56.09
sin H = \(\frac { 60 }{ 109 } \)
∠H = 33.3

Question 8.
Solve the right triangle. Round decimal answers to the nearest tenth.

Answer:
XY = 13.82, YZ = 6.69, ∠Y = 37.5

Explanation:
cos 52 = \(\frac { 8.5 }{ XY } \)
0.615 = \(\frac { 8.5 }{ XY } \)
XY = 13.82
sin 52 = \(\frac { YZ }{ XY } \)
0.788 = \(\frac { YZ }{ 8.5 } \)
YZ = 6.69
sin Y = \(\frac { 8.5 }{ 13.82 } \)
∠Y = 37.5

Question 9.
WHAT IF?
In Example 5, suppose another raked stage is 20 feet long from front to back with a total rise of 2 feet. Is the raked stage within your desired range?

Answer:
x = inverse sine of \(\frac { 2 }{ 20 } \)
x = 5.7°

Exercise 9.6 Solving Right Triangles

Question 1.
COMPLETE THE SENTENCE
To solve a right triangle means to find the measures of all its ________ and _______ .
Answer:

Question 2.
WRITING
Explain when you can use a trigonometric ratio to find a side length of a right triangle and when you can use the Pythagorean Theorem (Theorem 9.1 ).
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6. determine which of the two acute angles has the given trigonometric ratio.

Question 3.
The cosine of the angle is \(\frac{4}{5}\)

Answer:

Question 4.
The sine of the angle is \(\frac{5}{11}\)

Answer:
Sin(angle) = \(\frac { opposite }{ hypo } \)
sin A = \(\frac{5}{11}\)
The acute angle that has a sine of the angle is \(\frac{5}{11}\) is ∠A.

Question 5.
The sine of the angle is 0.95.

Answer:

Question 6.
The tangent of the angle is 1.5.

Answer:
tan(angle) = \(\frac { opposite }{ adjacent } \)
1.5 = \(\frac { 18 }{ 12 } \)
tan C = 1.5
The acute angle that has a tangent of the angle is 1.5 ∠C.

In Exercises 7 – 12, let ∠D be an acute angle. Use a calculator to approximate the measure of ∠D to the nearest tenth of a degree.

Question 7.
sin D = 0.75
Answer:

Question 8.
sin D = 0.19

Answer:
sin D = 0.19
∠D = inverse sine of 0.19
∠D = 10.9°

Question 9.
cos D = 0.33
Answer:

Question 10.
cos D = 0.64

Answer:
cos D = 0.64
∠D = inverse cos of 0.64
∠D = 50.2°

Question 11.
tan D = 0.28
Answer:

Question 12.
tan D = 0.72

Answer:
tan D = 0.72
∠D = inverse tan of 0.72
∠D = 35.8°

In Exercises 13 – 18. solve the right triangle. Round decimal answers to the nearest tenth.

Question 13.

Answer:

Question 14.

Answer:
ED = 2√65, ∠E = 59.3, ∠D = 29.7

Explanation:
c² = 8² + 14²
x² = 64 + 196
x² = 260
x = 2√65
sin E = \(\frac { 14 }{ 2√65 } \)
∠E = 59.3
sin D = \(\frac { 8 }{ 2√65 } \)
∠D = 29.7

Question 15.

Answer:

Question 16.

Answer:
HJ = 2√15, ∠G = 28.9, ∠J = 61

Explanation:
c² = a² + b²
16² = 14² + x²
x² = 256 – 196
x² = 60
x = 2√15
sin G = \(\frac { 2√15 }{ 16 } \)
∠G = 28.9
sin J = \(\frac { 14 }{ 16 } \)
∠J = 61

Question 17.

Answer:

Question 18.

Answer:
RT = 17.8, RS = 9.68, ∠T = 32.8

Explanation:
sin 57 = \(\frac { 15 }{ x } \)
0.838 = \(\frac { 15 }{ x } \)
x = 17.899
RT = 17.8
cos 57 = \(\frac { x }{ 17.8 } \)
0.544 = \(\frac { x }{ 17.8 } \)
x = 9.68
RS = 9.68
sin T = \(\frac { 9.68 }{ 17.8 } \)
∠T = 32.8

Question 19.
ERROR ANALYSIS
Describe and correct the error in using an inverse trigonometric ratio.

Answer:

Question 20.
PROBLEM SOLVING
In order to unload clay easily. the body of a dump truck must be elevated to at least 45° The body of a dump truck that is 14 feet long has been raised 8 feet. Will the clay pour out easily? Explain your reasoning.

Answer:
Angle of elevation: sin x = \(\frac { 8 }{ 14 } \)
x = inverse sine of \(\frac { 8 }{ 14 } \) = 34.9
The clay will not pour out easily.

Question 21.
PROBLEM SOLVING
You are standing on a footbridge that is 12 feet above a lake. You look down and see a duck in the water. The duck is 7 feet away from the footbridge. What is the angle of elevation from the duck to you

Answer:

Question 22.
HOW DO YOU SEE IT?
Write three expressions that can be used to approximate the measure of ∠A. Which expression would you choose? Explain your choice.

Answer:
Three expressions are ∠A = inverse tan of (\(\frac { 15 }{ 22 } \)) = 34.2°
∠A = inverse sine of (\(\frac { 15 }{ BA } \))
∠A = inverse cos of (\(\frac { 22 }{ BA } \))

Question 23.
MODELING WITH MATHEMATICS
The Uniform Federal Accessibility Standards specify that awheel chair ramp may not have an incline greater than 4.76. You want to build a ramp with a vertical rise of 8 inches. you want to minimize the horizontal distance taken up by the ramp. Draw a diagram showing the approximate dimensions of your ramp.
Answer:

Question 24.
MODELING WITH MATHEMATICS
The horizontal part of a step is called the tread. The vertical part is called the riser. The recommended riser – to – tread ratio is 7 inches : 11 inches.

a. Find the value of x for stairs built using the recommended riser-to-tread ratio.

Answer:

b. you want to build stairs that are less steep than the stairs in part (a). Give an example of a riser – to – tread ratio that you could use. Find the value of x for your stairs.
Answer:

Question 25.
USING TOOLS
Find the measure of ∠R without using a protractor. Justify your technique.

Answer:

Question 26.
MAKING AN ARGUMENT
Your friend claims that tan^-1x = \(\frac{1}{\tan x}\). Is your friend correct? Explain your reasoning.

Answer:
No
For example
tan^-1(√3) = 60
\(\frac{1}{\tan √3}\) = 33.1

USING STRUCTURE
In Exercises 27 and 28, solve each triangle.

Question 27.
∆JKM and ∆LKM

Answer:

Question 28.
∆TUS and ∆VTW

Answer:
TS = 8.2, UT = 7.3, ∠T = 28.6
TV = 13.2, TW = 9.6, ∠V = 46, ∠T = 42.84

Explanation:
tan 64 = \(\frac { TS }{ 4 } \)
TS = 2.05 x 4
TS = 8.2
sin 64 = \(\frac { UT }{ 8.2 } \)
0.898= \(\frac { UT }{ 8.2 } \)
UT = 7.3
sin T = \(\frac { 4 }{ 8.2 } \)
∠T = 28.6
TV = TS + SV
TV = 8.2 + 5 = 13.2
13.2² = TW² + 9²
TW² = 174.24 – 81
TW = 9.6
sin V = \(\frac { 9.6 }{ 13.2 } \)
∠V = 46
sin T = \(\frac { 9 }{ 13.2 } \)
∠T = 42.84

Question 29.
MATHEMATICAL CONNECTIONS
Write an expression that can be used to find the measure of the acute angle formed by each line and the x-axis. Then approximate the angle measure to the nearest tenth of a degree.
a. y = 3x
b. y = \(\frac{4}{3}\)x + 4
Answer:

Question 30.
THOUGHT PROVOKING
Simplify each expression. Justify your answer.
a. sin^-1 (sin x)

Answer:
sin^-1 (sin x) = x

b. tan(tan^-1 y)

Answer:
tan(tan^-1 y) = y

C. cos(cos^-1 z)

Answer:
cos(cos^-1 z) = z

Question 31.
REASONING
Explain why the expression sin^-1 (1.2) does not make sense.
Answer:

Question 32.
USING STRUCTURE
The perimeter of the rectangle ABCD is 16 centimeters. and the ratio of its width to its length is 1 : 3. Segment BD divides the rectangle into two congruent triangles. Find the side lengths and angle measures of these two triangles.

Answer:
The perimeter of the rectangle ABCD is 16 centimeters
2(l + b) = 16
l + b = 8
b : l = 1 : 3
4x = 8
x = 2
l = 6, b = 2
BD = √(6² + 2²) = √40 = 2√10
sin B = \(\frac { AD }{ BD } \) = \(\frac { 2 }{ 2√10 } \)
∠ABD = 18.4
∠CBD = 71.6
sin D = \(\frac { AB }{ BD } \) = \(\frac { 6 }{ 2√10 } \)
∠ADB = 71
∠CDB = 19

Maintaining Mathematical Proficiency

Solve the equation

Question 33.
\(\frac{12}{x}=\frac{3}{2}\)
Answer:

Question 34.
\(\frac{13}{9}=\frac{x}{18}\)

Answer:
\(\frac{13}{9}=\frac{x}{18}\)
x = 1.44 x 18
x = 26

Question 35.
\(\frac{x}{2.1}=\frac{4.1}{3.5}\)
Answer:

Question 36.
\(\frac{5.6}{12.7}=\frac{4.9}{x}\)

Answer:
\(\frac{5.6}{12.7}=\frac{4.9}{x}\)
0.44 = 4.9/x
x = 11.13

9.7 Law of Sines and Law of Cosines

Exploration 1

Discovering the Law of Sines

Work with a partner.

a. Copy and complete the table for the triangle shown. What can you conclude?


Answer:

b. Use dynamic geometry software to draw two other triangles. Copy and complete the table in part (a) for each triangle. Use your results to write a conjecture about the relationship between the sines of the angles and the lengths of the sides of a triangle.
USING TOOLS STRATEGICALLY
To be proficient in math, you need to use technology to compare predictions with data.
Answer:

Exploration 2

Discovering the Law of Cosines

Work with a partner:

a. Copy and complete the table for the triangle in Exploration 1 (a). What can you conclude?

Answer:

b. Use dynamic geometry software to draw two other triangles. Copy and complete the table in part (a) for each triangle. Use your results to write a conjecture about what you observe in the completed tables.
Answer:

Communicate Your Answer

Question 3.
What are the Law of Sines and the Law of Cosines?
Answer:

Question 4.
When would you use the Law of Sines to solve a triangle? When would you use the Law of Cosines to solve a triangle?
Answer:

Lesson 9.7 Law of Sines and Law of Cosines

Monitoring Progress

Use a calculator to find the trigonometric ratio. Round your answer to four decimal places.

Question 1.
tan 110°

Answer:
tan 110° = -2.7474

Question 2.
sin 97°

Answer:
sin 97° = 0.9925

Question 3.
cos 165°

Answer:
cos 165° = -0.9659

Find the area of ∆ABC with the given side lengths and included angle. Round your answer to the nearest tenth.

Question 4.
m ∠ B = 60°, a = 19, c = 14

Answer:
Area = 155.18

Explanation:
Area = \(\frac { 1 }{ 2 } \) ac sin B
= \(\frac { 1 }{ 2 } \) (19 x 14) sin 60°
= 133 x 0.866
= 155.18

Question 5.
m ∠ C = 29°, a = 38, b = 31

Answer:
Area = 282.72

Explanation:
Area = \(\frac { 1 }{ 2 } \) ab sin C
= \(\frac { 1 }{ 2 } \) (38 x 31) sin 29
= 598 x 0.48
= 282.72

Solve the triangle. Round decimal answers to the nearest tenth.

Question 6.

Answer:
∠C = 46.6, ∠B = 82.4, AC = 23.57

Explanation:
Using law of sines
\(\frac { c }{ sin C } \) = \(\frac { a }{ sin A } \)
\(\frac { 17 }{ sin C } \) = \(\frac { 18 }{ sin 51 } \)
\(\frac { 17 }{ sin C } \) = \(\frac { 18 }{ 0.77 } \)
sin C = 0.7274
∠C = 46.6
∠A + ∠B + ∠C  = 180
51 + 46.6 + ∠B = 180
∠B = 82.4
\(\frac { sin B }{ b } \) = \(\frac { sin A }{ a } \)
\(\frac { 0.99 }{ b } \) = \(\frac { 0.77 }{ 18 } \)
b = 23.57

Question 7.

Answer:
∠B = 31.3, ∠C  = 108.7, c = 23.6

Explanation:
\(\frac { sin B }{ b } \) = \(\frac { sin A }{ a } \)
\(\frac { sin B }{ 13 } \) = \(\frac { sin 40 }{ 16 } \)
sin B = \(\frac { .64 }{ 16 } \) x 13
sin B = 0.52
∠B = 31.3
∠A + ∠B + ∠C  = 180
40 + 31.3 + ∠C  = 180
∠C  = 108.7
\(\frac { c }{ sin C } \) = \(\frac { a }{ sin A } \)
\(\frac { c }{ sin 108.7 } \) = \(\frac {16 }{ sin 40 } \)
c = \(\frac {16 }{ .64 } \) x 0.947
c = 23.6

Solve the triangle. Round decimal answers to the nearest tenth.

Question 8.

Answer:
∠C = 66, a = 4.36, c = 8.27

Explanation:
∠A + ∠B + ∠C = 180
29 + 85 + ∠C = 180
∠C = 66
\(\frac { a }{ sin A } \) = \(\frac { b }{ sin B } \) = \(\frac { c }{ sin C } \)
\(\frac { a }{ sin 29 } \) = \(\frac { 9 }{ sin 85 } \)
\(\frac { a }{ 0.48 } \) = \(\frac { 9 }{ 0.99 } \)
a = 4.36
\(\frac { b }{ sin B } \) = \(\frac { c }{ sin C } \)
\(\frac { 9 }{ sin 85 } \) = \(\frac { c }{ sin 66 } \)
\(\frac { 9 }{ 0.99 } \) = \(\frac { c }{ 0.91 } \)
c = 8.27

Question 9.

Answer:
∠A = 29, b = 19.37, c = 20.41

Explanation:
∠A + ∠B + ∠C  = 180
∠A + 70 + 81 = 180
∠A = 29
\(\frac { a }{ sin A } \) = \(\frac { b }{ sin B } \) = \(\frac { c }{ sin C } \)
\(\frac { 10 }{ sin 29 } \) = \(\frac { b }{ sin 70 } \)
\(\frac { 10 }{ 0.48 } \) = \(\frac { b }{ 0.93 } \)
b = 19.37
\(\frac { a }{ sin A } \) = \(\frac { c }{ sin C } \)
\(\frac { 10 }{ sin 29 } \) = \(\frac { c }{ sin 81 } \)
\(\frac { 10 }{ 0.48 } \) = \(\frac { c }{ 0.98 } \)
c = 20.41

Question 10.
WHAT IF?
In Example 5, what would be the length of a bridge from the South Picnic Area to the East Picnic Area?

Answer:
The length of a bridge from the South Picnic Area to the East Picnic Area is 188 m.

Explanation:
\(\frac { a }{ sin A } \) = \(\frac { b }{ sin B } \) = \(\frac { c }{ sin C } \)
\(\frac { a }{ sin 71 } \) = \(\frac { 150 }{ sin 49 } \)
a = 188

Solve the triangle. Round decimal answers to the nearest tenth.

Question 11.

Answer:
b = 61.3, ∠A = 46, ∠C  = 46

Explanation:
b² = a² + c² − 2ac cos B
b² = 45² + 43² – 2(45)(43) cos 88
b² = 2025 + 1849 – 3870 x 0.03 = 3757.9
b = 61.3
\(\frac { sin A }{ a } \) = \(\frac { sin B }{ b } \)
\(\frac { sin A }{ 45 } \) = \(\frac { sin 88 }{ 61.3 } \)
sin A = 0.72
∠A = 46
∠A + ∠B + ∠C  = 180
46 + 88 + ∠C  = 180
∠C  = 46

Question 12.

Answer:
a = 41.1, ∠C  = 35.6, ∠B = 30.4

Explanation:
a² = b² + c² − 2bc cos A
a² = 23² + 26² – 2(23)(26) cos 114
a² = 529 + 676 – 1196 x -0.406
a² = 1690.5
a = 41.1
\(\frac { sin 114 }{ 41.1 } \) = \(\frac { sin B }{ 23 } \)
0.02 = \(\frac { sin B }{ 23 } \)
sin B = 0.507
∠B = 30.4
∠A + ∠B + ∠C  = 180
114 + 30.4 + ∠C  = 180
∠C  = 35.6

Question 13.

Answer:
∠A = 41.4, ∠B = 81.8, ∠C  = 56.8

Explanation:
a² = b² + c² − 2bc cos A
4² = 6² + 5² – 2(6)(5) cos A
16 = 36 + 25 – 60 cos A
-45 = – 60 cos A
cos A = 0.75
∠A = 41.4
\(\frac { sin 41.4 }{ 4 } \) = \(\frac { sin B }{ 6 } \)
0.165 = \(\frac { sin B }{ 6 } \)
sin B = 0.99
∠B = 81.8
∠A + ∠B + ∠C  = 180
41.4 + 81.8 + ∠C  = 180
∠C  = 56.8

Question 14.

Answer:
∠B = 81.8, ∠A = 58.6, ∠C  = 39.6

Explanation:
a² = b² + c² − 2bc cos A
23² = 27² + 16² – 2(27)(16) cos A
529 = 729 + 256 – 864 cos A
456 = 864 cos A
cos A = 0.52
∠A = 58.6
\(\frac { sin 58.6 }{ 23 } \) = \(\frac { sin B }{ 27 } \)
0.03 = \(\frac { sin B }{ 27 } \)
sin B = 0.99
∠B = 81.8
∠A + ∠B + ∠C  = 180
58.6 + 81.8 + ∠C  = 180
∠C  = 39.6

Exercise 9.7 Law of Sines and Law of Cosines

Vocabulary and Core Concept Check

Question 1.
WRITING
What type of triangle would you use the Law of Sines or the Law of Cosines to solve?
Answer:

Question 2.
VOCABULARY
What information do you need to use the Law of Sines?

Answer:

Monitoring progress and Modeling with Mathematics

In Exercises 3 – 8, use a calculator to find the trigonometric ratio, Round your answer to four decimal places.

Question 3.
sin 127°
Answer:

Question 4.
sin 98°

Answer:
sin 98° = 0.9902

Question 5.
cos 139°
Answer:

Question 6.
cos 108°

Answer:
cos 108° = -0.309

Question 7.
tan 165°
Answer:

Question 8.
tan 116°

Answer:
tan 116° = -2.0503

In Exercises 9 – 12, find the area of the triangle. Round your answer to the nearest tenth.

Question 9.

Answer:

Question 10.

Answer:
Area = \(\frac{1}{2}\)bc sin A
Area = \(\frac{1}{2}\)(28)(24) sin83
Area = 332.64

Question 11.

Answer:

Question 12.

Answer:
Area = \(\frac{1}{2}\)ab sin C
Area = \(\frac{1}{2}\)(15)(7) sin 96
Area = 51.9

In Exercises 13 – 18. solve the triangle. Round decimal answers to the nearest tenth.

Question 13.

Answer:

Question 14.

Answer:
∠B = 38.3, ∠A = 37.7, a = 15.7

Explanation:
\(\frac { sin B }{16 } \) = \(\frac { sin 104 }{ 25 } \)
sin B = 0.62
∠B = 38.3
∠A + ∠B + ∠C = 180
∠A + 38.3 + 104 = 180
∠A = 37.7
\(\frac { sin 37.7 }{ a } \) = \(\frac { sin 104 }{ 25 } \)
\(\frac { 0.61 }{ a } \) = 0.0388
a = 15.7

Question 15.

Answer:

Question 16.

Answer:
∠B = 65, b = 33.55, a = 24.4

Explanation:
∠A + ∠B + ∠C = 180
42 + 73 + ∠B = 180
∠B = 65
\(\frac { sin B }{ b } \) = \(\frac { sin C }{ c } \)
\(\frac { sin 65 }{ b } \) = \(\frac { sin 73 }{ 34 } \)
b = 33.55
\(\frac { sin A }{ a } \) = \(\frac { sin C }{ c } \)
\(\frac { sin 42 }{ a } \) = \(\frac { sin 73 }{ 34 } \)
a = 24.4

Question 17.

Answer:

Question 18.

Answer:
∠C = 90, b = 39.56, a = 17.6

Explanation:
∠A + ∠B + ∠C = 180
24 + 66 + ∠C = 180
∠C = 90
\(\frac { sin B }{ b } \) = \(\frac { sin C }{ c } \)
\(\frac { sin 66 }{ b } \) = \(\frac { sin 90 }{ 43 } \)
\(\frac { .91 }{ b } \) = 0.023
b = 39.56
\(\frac { sin A }{ a } \) = \(\frac { sin C }{ c } \)
\(\frac { sin 24 }{a } \) = \(\frac { sin 90 }{ 43 } \)
\(\frac { 0.406 }{a } \) = 0.023
a = 17.6

In Exercises 19 – 24, solve the triangle. Round decimal answers to the nearest tenth.

Question 19.

Answer:

Question 20.

Answer:
b = 29.9, ∠A = 26.1, ∠C = 15.07

Explanation:
b² = a² + c² – 2ac cos B
b² = 20² + 12² – 2(12)(20) cos 138
b² = 400 + 144 – 480 (-0.74)
b² = 899.2
b = 29.9
\(\frac { sin B }{ b } \) = \(\frac { sin A }{ a } \)
\(\frac { sin 138 }{ 29.9 } \) = \(\frac { sin A }{ 20 } \)
\(\frac { 0.66 }{ 29.9 } \) = \(\frac { sin A }{ 20 } \)
sin A = 0.44
∠A = 26.1
\(\frac { sin B }{ b } \) = \(\frac { sin C }{ c } \)
\(\frac { sin 138 }{ 29.9 } \) = \(\frac { sin C }{ 12 } \)
sin C = 0.26
∠C = 15.07

Question 21.

Answer:

Question 22.

Answer:
∠A = 107.3, ∠B = 51.6, ∠C = 21.1

Explanation:
b² = a² + c² – 2ac cos B
28² = 18² + 13² – 2(18)(13) cos B
784 = 324 + 169 – 468 cos B
291 = 468 cos B
cos B = 0.62
∠B = 51.6
\(\frac { sin C }{ c } \) = \(\frac { sin B }{ b } \)
\(\frac { sin C }{ 13 } \) = \(\frac { sin 51.6 }{ 28 } \)
sin C = 0.36
∠C = 21.1
∠A + ∠B + ∠C = 180
51.6 + 21.1 + ∠A = 180
∠A = 107.3

Question 23.

Answer:

Question 24.

Answer:
∠A = 23, ∠B = 132.1, ∠C = 24.9

Explanation:
b² = a² + c² – 2ac cos B
5² = 12² + 13² – 2(12)(13) cos B
25 = 144 + 169 – 312 cos B
288 = 312 cos B
cos B = 0.92
∠B = 23
\(\frac { sin C }{ c } \) = \(\frac { sin B }{ b } \)
\(\frac { sin C }{ 13 } \) = \(\frac { sin 23 }{ 5 } \)
sin C = 1.014
∠C = 24.9
∠A + ∠B + ∠C = 180
23 + 24.9 + ∠B = 180
∠B = 132.1

Question 25.
ERROR ANALYSIS
Describe and correct the error in finding m ∠ C.

Answer:

Question 26.
ERROR ANALYSIS
Describe and correct the error in finding m ∠ A in ∆ABC when a = 19, b = 21, and c = 11.

Answer:
a² = b² + c² – 2bc cos A
19² = 21² + 11² – 2(21)(11) cos A
361 = 441 + 121 – 462 cosA
201 = 462 cosA
cos A = 0.43
∠A = 64.5

COMPARING METHODS

In Exercise 27 – 32. tell whether you would use the Law of Sines, the Law of Cosines. or the Pythagorean Theorem (Theorem 9.1) and trigonometric ratios to solve the triangle with the given information. Explain your reasoning. Then solve the triangle.

Question 27.
m ∠ A = 72°, m ∠ B = 44°, b = 14
Answer:

Question 28.
m ∠ B = 98°, m ∠ C = 37°, a = 18

Answer:
∠A = 45, b = 25.38, c = 15.38

Explanation:
∠A + ∠B + ∠C = 180
∠A + 98 + 37 = 180
∠A = 45
\(\frac { sin B }{ b } \) = \(\frac { sin A }{ a } \)
\(\frac { sin 98 }{ b } \) = \(\frac { sin 45 }{ 18 } \)
\(\frac { 0.99 }{ b } \) = 0.039
b = 25.38
\(\frac { sin A }{ a } \) = \(\frac { sin C }{ c } \)
\(\frac { sin 45 }{ 18 } \) = \(\frac { sin 37 }{ c } \)
0.039 = \(\frac { sin 37 }{ c } \)
c = 15.38

Question 29.
m ∠ C = 65°, a = 12, b = 21
Answer:

Question 30.
m ∠ B = 90°, a = 15, c = 6

Answer:
b = 3√29, ∠A = 66.9, ∠C = 23.1

Explanation:
b² = a² + c²- 2ac cos B
b² = 15² + 6² – 2(15)(6) cos 90
= 225 + 36 – 180(0)
b² = 261
b = 3√29
\(\frac { sin B }{ b } \) = \(\frac { sin A }{ a } \)
\(\frac { sin 90 }{ 3√29 } \) = \(\frac { sin A }{ 15 } \)
sin A = 0.92
∠A = 66.9
∠A + ∠B + ∠C = 180
66.9 + 90 + ∠C = 180
∠C = 23.1

Question 31.
m ∠ C = 40°, b = 27, c = 36
Answer:

Question 32.
a = 34, b = 19, c = 27

Answer:
∠B = 33.9, ∠A = 78.5, ∠C = 67.6

Explanation:
b² = a² + c²- 2ac cos B
19² = 34² + 27²- 2(34)(27) cos B
361 = 1156 + 729 – 1836 cos B
cos B = 0.83
∠B = 33.9
\(\frac { sin 33.9 }{ 19 } \) = \(\frac { sin A }{ 34 } \)
sin A = 0.98
∠A = 78.5
∠A + ∠B + ∠C = 180
78.5 + 33.9 + ∠C = 180
∠C = 67.6

Question 33.
MODELING WITH MATHEMATICS
You and your friend are standing on the baseline of a basketball court. You bounce a basketball to your friend, as shown in the diagram. What is the distance between you and your friend?

Answer:

Question 34.
MODELING WITH MATHEMATICS
A zip line is constructed across a valley, as shown in the diagram. What is the width w of the valley?

Answer:
w = 92.5 ft

Explanation:
w² = 25² + 84² – 2(25)(84) cos 102
w² = 7681 – 4200 cos 102
w = 92.5 ft

Question 35.
MODELING WITH MATHEMATICS
You are on the observation deck of the Empire State Building looking at the Chrysler Building. When you turn 145° clockwise, you see the Statue of Liberty. You know that the Chrysler Building and the Empire Slate Building arc about 0.6 mile apart and that the Chrysler Building and the Statue of Liberty are about 5.6 miles apart. Estimate the distance between the Empire State Building and the Statue of Liberty.
Answer:

Question 36.
MODELING WITH MATHEMATICS
The Leaning Tower of Pisa in Italy has a height of 183 feet and is 4° off vertical. Find the horizontal distance d that the top of the tower is off vertical.

Answer:

Question 37.
MAKING AN ARGUMENT
Your friend says that the Law of Sines can be used to find JK. Your cousin says that the Law of Cosines can be used to find JK. Who is correct’? Explain your reasoning.

Answer:

Question 38.
REASONING
Use ∆XYZ

a. Can you use the Law of Sines to solve ∆XYZ ? Explain your reasoning.
Answer:

b. Can you use another method to solve ∆XYZ ? Explain your reasoning.
Answer:

Question 39.
MAKING AN ARGUMENT
Your friend calculates the area of the triangle using the formula A = \(\frac{1}{2}\)qr sin S and says that the area is approximately 208.6 square units. Is your friend correct? Explain your reasoning.

Answer:

Question 40.
MODELING WITH MATHEMATICS
You are fertilizing a triangular garden. One side of the garden is 62 feet long, and another side is 54 feet long. The angle opposite the 62-foot side is 58°.
a. Draw a diagram to represent this situation.
b. Use the Law of Sines to solve the triangle from part (a).
c. One bag of fertilizer covers an area of 200 square feet. How many bags of fertilizer will you need to cover the entire garden?

Answer:
C = 47.6, A = 74.4, a = 70.4
9 bags of fertilizer.

Question 41.
MODELING WITH MATHEMATICS
A golfer hits a drive 260 yards on a hole that is 400 yards long. The shot is 15° off target.

a. What is the distance x from the golfer’s ball to the hole?
b. Assume the golfer is able to hit the ball precisely the distance found in part (a). What is the maximum angle θ (theta) by which the ball can be off target in order to land no more than 10 yards fr0m the hole?
Answer:

Question 42.
COMPARING METHODS
A building is constructed on top of a cliff that is 300 meters high. A person standing on level ground below the cliff observes that the angle of elevation to the top of the building is 72° and the angle of elevation to the top of the cliff is 63°.
a. How far away is the person from the base of the cliff?

Answer:

b. Describe two different methods you can use to find the height of the building. Use one of these methods to find the building’s height.

Answer:
Consider △SYZ and evaluate d using tangent function
tan SYZ = \(\frac { 300 }{ d } \)
d = \(\frac { 300 }{ tan 63 } \)
d = 152.86
The person is standing 152.86 m away from the base of the cliff.
Consider △XYS and evaluate h + 300
tan XYZ = \(\frac { h + 300 }{ d } \)
h = 152.86 x tan 72 – 300
h = 170.45
The building is 170.45 m high.

Question 43.
MATHEMATICAL CONNECTIONS
Find the values of x and y.

Answer:

Question 44.
HOW DO YOU SEE IT?
Would you use the Law of Sines or the Law of Cosines to solve the triangle?
Answer:

Question 45.
REWRITING A FORMULA
A Simplify the Law of Cosines for when the given angle is a right angle.
Answer:

Question 46.
THOUGHT PROVOKING
Consider any triangle with side lengths of a, b, and c. Calculate the value of s, which is half the perimeter of the triangle. What measurement of the triangle is represented by \(\sqrt{s(s-a)(s-b)(s-c)} ?\)
Answer:

Question 47.
ANALYZING RELATIONSHIPS
The ambiguous case of the Law of Sines occurs when you are given the measure of one acute angle. the length of one adjacent side, and the length of the side opposite that angle, which is less than the length of the adjacent side. This results in two possible triangles. Using the given information, find two possible solutions for ∆ABC
Draw a diagram for each triangle.
(Hint: The inverse sine function gives only acute angle measures. so consider the acute angle and its supplement for ∠B.)

a. m ∠ A = 40°, a = 13, b = 16
b. m ∠ A = 21°, a = 17, b = 32
Answer:

Question 48.
ABSTRACT REASONING
Use the Law of Cosines to show that the measure of each angle of an equilateral triangle is 60°. Explain your reasoning.

Answer:
a² = b² + c²- 2bc cos A
a² = a² + a² – 2 aa cos A
a² = 2a² coas A
cos A = 1/2
∠A = 60

Question 49.
CRITICAL THINKING
An airplane flies 55° east of north from City A to City B. a distance of 470 miles. Another airplane flies 7° north of east from City A to City C. a distance of 890 miles. What is the distance between Cities B and C?
Answer:

Question 50.
REWRITING A FORMULA
Follow the steps to derive the formula for the area of a triangle.
Area = \(\frac{1}{2}\)ab sin C.

a. Draw the altitude from vertex B to \(\overline{A C}\). Label the altitude as h. Write a formula for the area of the triangle using h.
Answer:

b. Write an equation for sin C
Answer:

c. Use the results of parts (a) and (b) to write a formula for the area of a triangle that does not include h.
Answer:

Question 51.
PROVING A THEOREM
Follow the steps to use the formula for the area of a triangle to prove the Law of Sines (Theorem 9.9).

a. Use the derivation in Exercise 50 to explain how to derive the three related formulas for the area of a triangle.
Area = \(\frac{1}{2}\)bc sin A,
Area = \(\frac{1}{2}\)ac sin B,
Area = \(\frac{1}{2}\)ab sin C
b. why can you use the formulas in part (a) to write the following statement?
\(\frac{1}{2}\)bc sin A = \(\frac{1}{2}\)ac sin B = \(\frac{1}{2}\)ab sin C
c. Show how to rewrite the statement in part (b) to prove the Law of Sines. Justify each step.
Answer:

Question 52.
PROVING A THEOREM
Use the given information to complete the two – column proof of the Law of Cosines (Theorem 9.10).

Given \(\overline{B D}\) is an altitude of ∆ABC.
Prove a² = b² + c² – 2bc cos A

Statements	Reasons
1. \(\overline{B D}\) is an altitude of ∆ABC.	1. Given
2. ∆ADB and ∆CDB are right triangles.	2. _______________________
3. a2 = (b – x)2 + h2	3. _______________________
4. _______________________	4. Expand binomial.
5. x2 + h2 = c2	5. _______________________
6. _______________________	6. Substitution Property of Equality
7. cos A = \(\frac{x}{c}\)	7. _______________________
8. x = c cos A	8. _______________________
9. a2 = b2 + c2 – 2bc Cos A	9. _______________________

Answer:

Maintaining Mathematical Proficiency

Find the radius and diameter of the circle.

Question 53.

Answer:

Question 54.

Answer:
The radius is 10 in and the diameter is 20 in.

Question 55.

Answer:

Question 56.

Answer:
The radius is 50 in and the diameter is 100 in.

Right Triangles and Trigonometry Review

9.1 The Pythagorean Theorem

Find the value of x. Then tell whether the side lengths form a Pythagorean triple.

Question 1.

Answer:
x = 2√34
The sides will not form a Pythagorean triple.

Explanation:
x² = 6² + 10²
x²= 36 + 100
x = 2√34

Question 2.

Answer:
x = 12
The sides form a Pythagorean triple.

Explanation:
20² = 16²+ x²
400 = 256 + x²
x² = 144
x = 12

Question 3.

Answer:
x = 2√30
The sides will not form a Pythagorean triple.

Explanation:
13² = 7² + x²
169 = 49 + x²
x = 2√30

Verify that the segments lengths form a triangle. Is the triangle acute, right, or obtuse?

Question 4.
6, 8, and 9

Answer:
9² = 81
6² + 8² = 36 + 64 = 100
9² < 6² + 8²
So, the triangle is acute

Question 5.
10, 2√2, and 6√3

Answer:
10² = 100
(2√2)² + (6√3)² = 8 + 108 = 116
So, the triangle is acute.

Question 6.
13, 18 and 3√55

Answer:
18² = 324
13² + (3√55)² = 169 + 495 = 664
So, the triangle is acute.

9.2 Special Right Triangles

Find the value of x. Write your answer in simplest form.

Question 7.

Answer:
hypotenuse = leg • √2
x = 6√2

Question 8.

Answer:
longer leg = shorter leg • √3
14 = x • √3
x = 8.08

Question 9.

Answer:
longer leg = shorter leg • √3
x = 8√3 • √3
x = 24

9.3 Similar Right Triangles

Identify the similar triangles. Then find the value of x.

Question 10.

Answer:
9 = √(6x)
81 = 6x
x = \(\frac { 27 }{ 2 } \)

Question 11.

Answer:
x = √(6 x 4)
x = 2√6

Question 12.

Answer:
\(\frac { RP }{ RQ } \) = \(\frac { SP }{ RS } \)
\(\frac { 9 }{ x } \) = \(\frac { 6 }{ 3 } \)
x = 3.5

Question 13.

Answer:
\(\frac { SU }{ ST } \) = \(\frac { SV }{ VU } \)
\(\frac { 16 }{ 20 } \) = \(\frac { x }{ (x – 16) } \)
4(x – 16) = 5x
x = 16

Find the geometric mean of the two numbers.

Question 14.
9 and 25

Answer:
mean = √(9 x 25)
= 15

Question 15.
36 and 48

Answer:
mean = √(36 x 48)
= 24√3

Question 16.
12 and 42

Answer:
mean = √(12 x 42)
= 6√14

9.4 The Tangent Ratio

Find the tangents of the acute angles in the right triangle. Write each answer as a fraction and as a decimal rounded to four decimal places.

Question 17.

Answer:
tan J = \(\frac { Opposite side }{ Adjacent side } \)
tan J = \(\frac { LK }{ JK } \) = \(\frac { 11 }{ 60 } \)
tan L = \(\frac { JK }{ LK } \) = \(\frac { 60 }{ 11 } \)

Question 18.

Answer:
tan P = \(\frac { MN }{ MP } \) = \(\frac { 35 }{ 12 } \)
tan N = \(\frac { MP }{ MN } \) = \(\frac { 12 }{ 35 } \)

Question 19.

Answer:
tan A = \(\frac { BC }{ AC } \) = \(\frac { 7 }{ 4√2 } \)
tan B = \(\frac { AC }{ BC } \) = \(\frac { 4√2 }{ 7 } \)

Find the value of x. Round your answer to the nearest tenth.

Question 20.

Answer:
tan 54 = \(\frac { x }{ 32 } \)
1.37 = \(\frac { x }{ 32 } \)
x = 43.8

Question 21.

Answer:
tan 25 = \(\frac { x }{ 20 } \)
0.46 x 20 = x
x = 9.2

Question 22.

Answer:
tan 38 = \(\frac { 10 }{ x } \)
x = 12.82

Question 23.
The angle between the bottom of a fence and the top of a tree is 75°. The tree is 4 let from the fence. How tall is the tree? Round your answer to the nearest foot.

Answer:
tan 75 = \(\frac { x }{ 4 } \)
x = 14.92

9.5 The Sine and Cosine Ratios

Find sin X, sin Z, cos X, and cos Z. Write each answer as a fraction and as a decimal rounded to four decimal places.

Question 24.

Answer:
sin X = \(\frac { 3 }{ 5 } \)
sin Z = \(\frac { 4 }{ 5 } \)
cos X = \(\frac { 4 }{ 5 } \)
cos Z = \(\frac { 3 }{ 5 } \)

Question 25.

Answer:
sin X = \(\frac { 7 }{ √149 } \)
sin Z = \(\frac { 10 }{ √149 } \)
cos X = \(\frac { 10 }{ √149 } \)
cos Z = \(\frac { 7 }{ √149 } \)

Question 26.

Answer:
sin X = \(\frac { 55 }{ 73 } \)
sin Z = \(\frac { 48 }{ 73 } \)
cos X = \(\frac { 48 }{ 73 } \)
cos Z = \(\frac { 55 }{ 73 } \)

Find the value of each variable using sine and cosine. Round your answers to the nearest tenth.

Question 27.

Answer:
sin 23 = \(\frac { t }{ 34 } \)
t = 13.26
cos 23 = \(\frac { s }{ 34 } \)
s = 31.28

Question 28.

Answer:
sin 36 = \(\frac { s }{ 5 } \)
s = 2.9
cos 36 = \(\frac { r }{ 5 } \)
r = 4

Question 29.

Answer:
sin 70 = \(\frac { v }{ 10 } \)
v = 9.39
cos 70 = \(\frac { w }{ 10 } \)
w = 3.42

Question 30.
Write sin 72° in terms of cosine.

Answer:
sin 72 = cos(90 – 72)
= cos 18 = 0.95

Question 31.
Write cos 29° in terms of sine.

Answer:
sin 29 = cos(90 – 29)
= cos 61 = 0.48

9.6 Solving Right Triangles

Let ∠Q be an acute angle. Use a calculator to approximate the measure of ∠Q to the nearest tenth of a degree.

Question 32.
cos Q = 0.32

Answer:
cos Q = 0.32
∠Q = inverse cos of .32
∠Q = 71.3

Question 33.
sin Q = 0.91

Answer:
sin Q = 0.91
∠Q = inverse sin of 0.91
∠Q = 65.5

Question 34.
tan Q = 0.04

Answer:
tan Q = 0.04
∠Q = inverse tan of 0.04
∠Q = 2.29

Solve the right triangle. Round decimal answers to the nearest tenth.

Question 35.

Answer:
a = 5√5, ∠A = 47.7, ∠B = 42.3

Explanation:
c² = a² + b²
15² = a² + 10²
a² = 125
a = 5√5
sin A = \(\frac { 5√5 }{ 15 } \) = 0.74
∠A = 47.7
∠A + ∠B + ∠C = 180
47.7 + ∠B + 90 = 180
∠B = 42.3

Question 36.

Answer:
NL = 7.59, ∠L = 53, ML = 4.55

Explanation:
cos 37 = \(\frac { 6 }{ NL } \)
NL = 7.59
∠N + ∠M + ∠L = 180
37 + 90 + ∠L = 180
∠L = 53
sin 37 = \(\frac { ML }{ 7.59 } \)
ML = 4.55

Question 37.

Answer:
XY = 17.34, ∠X = 46, ∠Z = 44

Explanation:
c² = a² + b²
25² = 18²+ b²
b² = 301
b = 17.34
sin X = \(\frac { 18 }{ 25 } \)
∠X = 46
sum of angles = 180
46 + 90 + ∠Z = 180
∠Z = 44

9.7 Law of Sines and Law of Cosines

Find the area of ∆ABC with the given side lengths and included angle.

Question 38.
m ∠ B = 124°, a = 9, c = 11

Answer:
Area = \(\frac { 1 }{ 2 } \) ac sin B
= \(\frac { 1 }{ 2 } \) (9 x 11) sin 124
= 40.59

Question 39.
m ∠ A = 68°, b = 13, c = 7

Answer:
Area = \(\frac { 1 }{ 2 } \) bc sin A
= \(\frac { 1 }{ 2 } \) (13 x 7) sin 68
= 41.86

Question 40.
m ∠ C = 79°, a = 25 b = 17

Answer:
Area = \(\frac { 1 }{ 2 } \) ab sin C
= \(\frac { 1 }{ 2 } \) (25 x 17) sin 79
= 208.25

Solve ∆ABC. Round decimal answers to the nearest tenth.

Question 41.
m ∠ A = 112°, a = 9, b = 4

Answer:
∠B = 24, ∠C = 44, c = 6.76

Explanation:
\(\frac { sin B }{ b } \) = \(\frac { sin A }{ a } \)
\(\frac { sin B }{ 4 } \) = \(\frac { sin 112 }{ 9 } \)
sin B = 0.408
∠B = 24
∠A + ∠B + ∠C = 180
112 + 24 + ∠C = 180
∠C = 44
\(\frac { sin 112 }{ 9 } \) = \(\frac { sin 44 }{ c } \)
c = 6.76

Question 42.
m ∠ 4 = 28°, m ∠ B = 64°, c = 55

Answer:
∠C = 88, b = 49.4, a = 25.5

Explanation:
∠A + ∠B + ∠C = 180
28 + 64 + ∠C = 180
∠C = 88
\(\frac { sin B }{ b } \) = \(\frac { sin C }{ c } \)
\(\frac { sin 64 }{ b } \) = \(\frac { sin 88 }{ 55 } \)
\(\frac { sin 64 }{ b } \) = 0.018
b = 49.4
\(\frac { sin 28 }{ a } \) = \(\frac { sin 88 }{ 55 } \)
a = 25.5

Question 43.
m ∠ C = 48°, b = 20, c = 28

Answer:
∠B = 31.3, ∠A = 100.7, a = 37.6

Explanation:
\(\frac { sin B }{ b } \) = \(\frac { sin C }{ c } \)
\(\frac { sin B }{ 20 } \) = \(\frac { sin 48 }{ 28 } \)
\(\frac { sin B }{ 20 } \) = 0.026
sin B = 0.52
∠B = 31.3
∠A + ∠B + ∠C = 180
∠A + 31.3 + 48 = 180
∠A = 100.7
\(\frac { sin 100.7 }{ a } \) = \(\frac { sin 48 }{ 28 } \)
a = 37.6

Question 44.
m ∠ B = 25°, a = 8, c = 3

Answer:
b = 5.45, ∠A = 37.5, ∠C = 117.5

Explanation:
b² = a² + c²- 2ac cos B
b² = 8² + 3² – 2(8 x 3) cos 25 = 73 – 43.2 = 29.8
b = 5.45
\(\frac { sin A }{ 8 } \) = \(\frac { sin 25 }{ 5.45 } \)
sin A = 0.61
∠A = 37.5
∠A + ∠B + ∠C = 180
37.5 + 25 + ∠C = 180
∠C = 117.5

Question 45.
m ∠ B = 102°, m ∠ C = 43°, b = 21

Answer:
∠A = 35, c = 14.72, a = 12.3

Explanation:
∠A + ∠B + ∠C = 180
∠A + 102 + 43 = 180
∠A = 35
\(\frac { sin 102 }{ 21 } \) = \(\frac { sin 43 }{ c } \)
0.046 = \(\frac { sin 43 }{ c } \)
c = 14.72
\(\frac { sin 102 }{ 21 } \) = \(\frac { sin 35 }{ a } \)
0.046 = \(\frac { sin 35 }{ a } \)
a = 12.3

Question 46.
a = 10, b = 3, c = 12

Answer:
∠B = 11.7, ∠C = 125.19, ∠A = 43.11

Explanation:
b² = a² + c²- 2ac cos B
3² = 10² + 12²- 2(10 x 12) cos B
9 = 100 + 144 – 240 cos B
cos B = 0.979
∠B = 11.7
a² = b² + c²- 2bc cos A
100 = 9 + 144 – 72 cos A
cos A = 0.73
∠A = 43.11
∠A + ∠B + ∠C = 180
43.11 + 11.7 + ∠C = 180
∠C = 125.19

Right Triangles and Trigonometry Test

Find the value of each variable. Round your answers to the nearest tenth.

Question 1.

Answer:
sin 25 = \(\frac { t }{ 18 } \)
t = 7.5
cos 25 = \(\frac { s }{ 18 } \)
s = 16.2

Question 2.

Answer:
sin 22 = \(\frac { 6 }{ x } \)
x = 16.21
cos 22 = \(\frac { y }{ 16.21 } \)
y = 14.91

Question 3.

Answer:
tan 40 = \(\frac { k }{ 10 } \)
k = 8.3
cos 40 = \(\frac { 10 }{ j } \)
j = 13.15

Verity that the segment lengths form a triangle. Is the triangle acute, right, or obtuse?

Question 4.
16, 30, and 34

Answer:
34²= 16² + 30²
So, the triangle is a right-angled triangle.

Question 5.
4, √67, and 9

Answer:
9² = 81
4² + (√67)² = 83
So the triangle is acute

Question 6.
√5. 5. and 5.5

Answer:
5.5² = 30.25
√5² + 5² = 30
So the triangle is obtuse

Solve ∆ABC. Round decimal answers to the nearest tenth.

Question 7.

Answer:
c = 12.08, ∠A = 24.22, ∠C = 65.78

Explanation:
tan A = \(\frac { 5 }{ 11 } \)
∠A = 24.22
c² = 11²+ 5²
c = 12.08
24.22 + 90 + ∠C = 180
∠C = 65.78

Question 8.

Answer:
∠B = 35.4, ∠C = 71.6, c = 17.9

Explanation:
\(\frac { sin 73 }{ 18 } \) = \(\frac { sin B }{ 11 } \)
sin B = 0.58
∠B = 35.4
73 + 35.4 + ∠C = 180
∠C = 71.6
\(\frac { sin 73 }{ 18 } \) = \(\frac { sin 71.6 }{ c } \)
c = 17.9

Question 9.

Answer:
BC = 4.54, ∠B = 59.3, ∠A = 30.7

Explanation:
9.2² = 8² + x²
x = 4.54
\(\frac { sin 90 }{ 9.2 } \) = \(\frac { sin B }{ 8 } \)
sin B = 0.86
∠B = 59.3
∠A + 59.3 + 90 = 180
∠A = 30.7

Question 10.
m ∠ A = 103°, b = 12, c = 24

Answer:
a = 29, ∠B = 53.5

Explanation:
a² = b² + c²- 2bc cos A
a² = 144 + 24² – 2(12 x 24) cos 103
a = 29
\(\frac { sin B }{ 12 } \) = \(\frac { sin 103 }{ 29 } \)
∠B = 23.5
∠C = 180 – (103 + 23.5) = 53.5

Question 11.
m ∠ A = 26°, m ∠ C = 35°, b = 13

Answer:
∠B = 119, a = 6.42, c = 8.5

Explanation:
∠B + 26 + 35 = 180
∠B = 119
\(\frac { sin 119 }{ 13 } \) = \(\frac { sin 26 }{ a } \)
a = 6.42
\(\frac { sin 119 }{ 13 } \) = \(\frac { sin 35 }{ c } \)
c = 8.5

Question 12.
a = 38, b = 31, c = 35

Answer:
∠B=50.2, ∠C = 59.8, ∠A = 70

Explanation:
b² = a² + c²- 2ac cos B
31² = 38²+ 35²- 2(35 x 38) cos B
cos B = 0.64
∠B=50.2
a² = b² + c²- 2bc cos A
38² = 31²+ 35²- 2(31 x 35) cos A
cos A = 0.341
∠A = 70
∠C = 59.8

Question 13.
Write cos 53° in terms of sine.

Answer:
cos 53° = sin (90 – 53) = sin 37

Find the value of each variable. Write your answers in simplest form.

Question 14.

Answer:
sin 45 = \(\frac { 16 }{ q } \)
q = 22.6
cos 45 = \(\frac { r }{ q } \)
r = 16

Question 15.

Answer:

Question 16.

Answer:
sin 30 = \(\frac { f }{ 9.2 } \)
f = 4.6
cos 30 = \(\frac { 8 }{ h } \)
h = 9.2

Question 17.
In ∆QRS, m ∠ R = 57°, q = 9, and s = 5. Find the area of ∆QRS.

Answer:
Area = \(\frac { 1 }{ 2 } \) qs sin R
= \(\frac { 1 }{ 2 } \) (9 x 5) sin 57 = 18.675

Question 18.
You are given the measures of both acute angles of a right triangle. Can you determine the side lengths? Explain.

Answer:
No.

Question 19.
You are at a parade looking up at a large balloon floating directly above the street. You are 60 feet from a point on the street directly beneath the balloon. To see the top of the balloon, you look up at an angle of 53°. To see the bottom of the balloon, you look up at an angle of 29°. Estimate the height h of the balloon.

Answer:

Question 20.
You warn to take a picture of a statue on Easter Island, called a moai. The moai is about 13 feet tall. Your camera is on a tripod that is 5 feet tall. The vertical viewing angle of your camera is set at 90°. How far from the moai should you stand so that the entire height of the moai is perfectly framed in the photo?

Answer:

Right Triangles and Trigonometry Cummulative Assessment

Question 1.
The size of a laptop screen is measured by the length of its diagonal. you Want to purchase a laptop with the largest screen possible. Which laptop should you buy?
(A)

(B)

(C)

(D)

Answer:
(B)

Explanation:
(a) d = √9² + 12² = 15
(b) d = √11.25² + 20² = 22.94
(c) d = √12² + 6.75² = 13.76
(d) d = √8² + 6² = 10

Question 2.
In ∆PQR and ∆SQT, S is between P and Q, T is between R and Q, and \(\) What must be true about \(\overline{S T}\) and \(\overline{P R}\)? Select all that apply.
\(\overline{S T}\) ⊥ \(\overline{P R}\)      \(\overline{S T}\) || \(\overline{P R}\)    ST = PR        ST = \(\frac{1}{2}\)PR
Answer:

Question 3.
In the diagram, ∆JKL ~ ∆QRS. Choose the symbol that makes each statement true.

<      =       >
sin J ___________ sin Q                 sin L ___________ cos J                   cos L ___________ tan Q
cos S ___________ cos                  J cos J ___________ sin S                 tan J ___________ tan Q
tan L ___________ tan Q               tan S ___________ cos Q                sin Q ___________ cos L
Answer:
sin J = sin Q                 sin L = cos J                   cos L = tan Q
cos S > cos J                  cos J > sin S                 tan J = tan Q
tan L < tan Q               tan S > cos Q                sin Q = cos L

Question 4.
A surveyor makes the measurements shown. What is the width of the river.

Answer:
tan 34 = \(\frac { AB }{ 84 } \)
AB = 56.28

Question 5.
Create as many true equations as possible.

____________ = ______________

sin X            cos X           tan x           \(\frac{X Y}{X Z}\)           \(\frac{Y Z}{X Z}\)

Sin Z            cos Z           tan Z           \(\frac{X Y}{Y Z}\)           \(\frac{Y Z}{X Y}\)

Answer:
sin X = \(\frac{Y Z}{X Z}\) = cos Z
cos X = \(\frac{X Y}{X Z}\) = sin Z
tan x = \(\frac{Y Z}{X Y}\)
tan Z = \(\frac{X Y}{Y Z}\)

Question 6.
Prove that quadrilateral DEFG is a kite.
Given \(\overline{H E} \cong \overline{H G}\), \(\overline{E G}\) ⊥ \(\overline{D F}\)
Prove \(\overline{F E} \cong \overline{F G}\), \(\overline{D E} \cong \overline{D G}\)

Answer:

Question 7.
What are the coordinates of the vertices of the image of ∆QRS after the composition of transformations show?

(A) Q’ (1, 2), R'(5, 4), S'(4, -1)
(B) Q'(- 1, – 2), R’ (- 5, – 4), S’ (- 4, 1)
(C) Q'(3, – 2), R’ (- 1, – 4), S’ (0, 1)
(D) Q’ (-2, 1), R'(- 4, 5), S'(1, 4)
Answer:

Question 8.
The Red Pyramid in Egypt has a square base. Each side of the base measures 722 feet. The height of the pyramid is 343 fee.

a. Use the side length of the base, the height of the pyramid, and the Pythagorean Theorem to find the slant height, AB, of the pyramid.

Answer:
343² = h² + 722²
h = 635.3

b. Find AC.
Answer:
AC = 343

c. Name three possible ways of finding m ∠ 1. Then, find m ∠ 1.
Answer:
Three possible ways are sin 1, cos 1 and tan 1
tan 1 = \(\frac { 722 }{ 635.3 } \)
∠1 = 48

Big Ideas Math Book Geometry Chapter 10 Circles Answers are provided here. Students who have been looking for the BIM Geometry Chapter 10 Circles Answers can read the following sections. The high school students can find a direct link to download Big Ideas Math Geometry Answers Chapter 10 Circles pdf for free of cost. With the help of this answer key, you can prepare well for the exam.

Big Ideas Math Book Geometry Answer Key Chapter 10 Circles

The different chapters included in Big Ideas Math Geometry Solutions are Lines and Segments That Intersect Circles, Finding Arc Measures, Inscribed Angles and Polygons, Angle Relationships in Circles, Segment Relationships in Circles, Circles in the Coordinate Plane, and Using Chords. Students have to practise all the questions from Big Ideas Math Textbook Geometry Chapter 10 Circles.

This Big Ideas Math Book Geometry Answer Key Chapter 10 Circles helps the students while doing the assignments. Get the solutions for all the questions through the quick links provided in the following sections. Test your skills through performance task, chapter review, and maintaining mathematical proficiency.

Circles Maintaining Mathematical Proficiency

Find the Product.

Question 1.
(x + 7) (x + 4)

Answer:
(x + 7) (x + 4) = x² + 14x + 28

Explanation:
(x + 7) (x + 4) = x(x + 7) + 7(x + 4)
= x² + 7x + 7x + 28
= x² + 14x + 28

Question 2.
(a + 1) (a – 5)

Answer:
(a + 1) (a – 5) = a² – 4a – 5

Explanation:
(a + 1) (a – 5) = a(a – 5) + 1(a – 5)
= a² – 5a + a – 5
= a² – 4a – 5

Question 3.
(q – 9) (3q – 4)

Answer:
(q – 9) (3q – 4) = 3q² – 31q + 32

Explanation:
(q – 9) (3q – 4) = q(3q – 4) – 9(3q – 4)Exercise 10.3 Using Chords

= 3q² – 4q – 27q + 32
= 3q² – 31q + 32

Question 4.
(2v – 7) (5v + 1)

Answer:
(2v – 7) (5v + 1) = 10v² – 33v – 7

Explanation:
(2v – 7) (5v + 1) = 2v(5v + 1)- 7(5v + 1)
= 10v² + 2v – 35v – 7
= 10v² – 33v – 7

Question 5.
(4h + 3) (2 + h)

Answer:
(4h + 3) (2 + h) = 4h² + 11h + 6

Explanation:
(4h + 3) (2 + h) = 4h(2 + h) + 3(2 + h)
= 8h + 4h² + 6 + 3h
= 4h² + 11h + 6

Question 6.
(8 – 6b) (5 – 3b)

Answer:
(8 – 6b) (5 – 3b) = 18b² – 54b + 40

Explanation:
(8 – 6b) (5 – 3b) = 8(5 – 3b) – 6b(5 – 3b)
= 40 – 24b – 30b + 18b²
= 18b² – 54b + 40

Solve the equation by completing the square. Round your answer to the nearest hundredth, if necessary.

Question 7.
x² – 2x = 5

Answer:
The solutions are x = √6 + 1, x = 1 – √6

Explanation:
x² – 2x = 5
x² – 2x + 1² = 5 + 1²
(x – 1)² = 6
x – 1 = ±√6
x = ±√6 + 1
The solutions are x = √6 + 1, x = -√6 + 1

Question 8.
r² + 10r = -7

Answer:
The solutions are r = √18 – 5, r = 5 – √18

Explanation:
r² + 10r = -7
r² + 10r + 5² = -7 + 5²
(r + 5)² = -7 + 25 = 18
r + 5 = ±√18
r = ±√18 – 5
The solutions are r = √18 – 5, r = 5 – √18

Question 9.
w² – 8w = 9

Answer:
The solutions are w = 9, w = -1

Explanation:
w² – 8w = 9
w² – 8w + 4² = 9 + 4²
(w – 4)² = 9 + 16 = 25
w – 4 = ±5
w = 5 + 4, w = -5 + 4
w = 9, w = -1
The solutions are w = 9, w = -1

Question 10.
p² + 10p – 4 = 0

Answer:
The solutions are p = √29 – 5, p = 5 – √29

Explanation:
p² + 10p = 4
p² + 10p + 5² = 4 + 5²
(p + 5)² = 4 + 25
(p + 5)² = 29
p + 5 = ±√29
p = ±√29 – 5
The solutions are p = √29 – 5, p = 5 – √29

Question 11.
k² – 4k – 7 = 0

Answer:
The solutions are k = √11 + 2, k = 2 – √11

Explanation:
k² – 4k= 7
k² – 4k + 2² = 7 + 4
(k – 2)² = 11
k – 2 = ±√11
k = √11 + 2, k = 2 – √11
The solutions are k = √11 + 2, k = 2 – √11

Question 12.
– z² + 2z = 1

Answer:
The solutions are z = 1

Explanation:
-z² + 2z = 1
z² – 2z = -1
z² – 2z + 1 = -1 + 1
(z – 1)² = 0
z = 1
The solutions are z = 1

Question 13.
ABSTRACT REASONING
write an expression that represents the product of two consecutive positive odd integers. Explain your reasoning.

Answer:
Let us take two consecutive odd integers are x and (x + 2)
The product of two consecutive odd integers is x • (x + 2)
x(x + 2) = x² + 2x

Circles Mathematical Practices

Monitoring progress

Let ⊙A, ⊙B, and ⊙C consist of points that are 3 units from the centers.

Question 1.
Draw ⊙C so that it passes through points A and B in the figure at the right. Explain your reasoning.

Answer:

Question 2.
Draw ⊙A, ⊙B, and OC so that each is tangent to the other two. Draw a larger circle, ⊙D, that is tangent to each of the other three circles. Is the distance from point D to a point on ⊙D less than, greater than, or equal to 6? Explain.

Answer:

10.1 Lines and Segments that Intersect Circles

Exploration 1

Lines and Line Segments That Intersect Circles

Work with a partner: The drawing at the right shows five lines or segments that intersect a circle. Use the relationships shown to write a definition for each type of line or segment. Then use the Internet or some other resource to verify your definitions.
Chord: _________________
Secant: _________________
Tangent: _________________
Radius: _________________
Diameter: _________________

Answer:
Chord: A chord of a circle is a straight line segment whose endpoints both lie on a circular arc.
Secant: A straight line that intersects a circle in two points is called a secant line.
Tangent: Tangent line is a line that intersects a curved line at exactly one point.
Radius: It is the distance from the centre of the circle to any point on the circle.
Diameter: It the straight that joins two points on the circle and passes through the centre of the circle.

Exploration 2

Using String to Draw a Circle

Work with a partner: Use two pencils, a piece of string, and a piece of paper.

a. Tie the two ends of the piece of string loosely around the two pencils.
Answer:

b. Anchor one pencil of the paper at the center of the circle. Use the other pencil to draw a circle around the anchor point while using slight pressure to keep the string taut. Do not let the string wind around either pencil.

Answer:

c. Explain how the distance between the two pencil points as you draw the circle is related to two of the lines or line segments you defined in Exploration 1.
REASONING ABSTRACTLY
To be proficient in math, you need to know and flexibly use different properties of operations and objects.
Answer:

Communicate Your Answer

Question 3.
What are the definitions of the lines and segments that intersect a circle?
Answer:

Question 4.
Of the five types of lines and segments in Exploration 1, which one is a subset of another? Explain.
Answer:

Question 5.
Explain how to draw a circle with a diameter of 8 inches.
Answer:

Lesson 10.1 Lines and Segments that Intersect Circles

Monitoring progress

Question 1.
In Example 1, What word best describes \(\overline{A G}\)? \(\overline{C B}\)?

Answer:
\(\overline{A G}\) is secant because it is a line that intersects the circle at two points.
\(\overline{C B}\) is the radius as it is the distance from the centre to the point of a circle.

Question 2.
In Example 1, name a tangent and a tangent segment.

Answer:
\(\overline{D E}\) is the tangent of the circle
\(\overline{D E}\) is the tangent segment of the circle.

Tell how many common tangents the circles have and draw them. State whether the tangents are external tangents or internal tangents.

Question 3.

Answer:
4 tangents.

A tangent is a line segment that intersects the circle at exactly one point. Internal tangents are the lines that intersect the segments joining the centres of two circles. External tangents are the lines that do not cross the segment joining the centres of the circles.
Blue lines represent the external tangents and red lines represent the internal tangents.

Question 4.

Answer:
One tangent.

One external tangent.

Question 5.

Answer:
No tangent.
It is not possible to draw a common tangent for this type of circle.

Question 6.
Is \(\overline{D E}\) tangent to ⊙C?

Answer:
Use the converse of Pythagorean theorem i.e 2² = 3² + 4²
4 = 9 + 16
By the tangent line to the circle theorem, \(\overline{D E}\) is not a tangent to ⊙C

Question 7.
\(\overline{S T}\) is tangent to ⊙Q.
Find the radius of ⊙Q.

Answer:
The radius of ⊙Q is 7 units.

Explanation:
By using the Pythagorean theorem
(18 + r)² = r² + 24²
324 + 36r + r² = r² + 576
36r = 576 – 324
36r = 252
r = 7

Question 8.
Points M and N are points of tangency. Find the value(s) of x.

Answer:
The values of x are 3 or -3.

Explanation:
x² = 9
x = ±3

Exercise 10.1 Lines and Segments that Intersect Circles

Vocabulary and Core Concept Check

Question 1.
WRITING
How are chords and secants alike? How are they different?
Answer:

Question 2.
WRITING
Explain how you can determine from the context whether the words radius and diameter are referring to segments or lengths.

Answer:
Radius and diameter are the lengths of the line segments that pass through the centre of a circle. Radius is half of the diameter.

Question 3.
COMPLETE THE SENTENCE
Coplanar circles that have a common center are called ____________ .
Answer:

Question 4.
WHICH ONE DOESNT BELONG?
Which segment does not belong with the other three? Explain your reasoning.
chord radius tangent diameter

Answer:
A chord, a radius and a diamter are segments and they intersect a circle in two points. A tangent is a line that intersects a circle at one point.

Monitoring Progress and Modeling with Mathematics

In Exercises 5 – 10, use the diagram.

Question 5.
Name the circle.
Answer:

Question 6.
Name two radii.

Answer:
The name of the two radii is CD and AC.

Question 7.
Name two chords.
Answer:

Question 8.
Name a diameter.

Answer:
The name of diameter is AD

Question 9.
Name a secant.
Answer:

Question 10.
Name a tangent and a point of tangency

Answer:
GE is the tangent and F is the point of tangency.

In Exercises 11 – 14, copy the diagram. Tell how many common tangents the circles have and draw them.

Question 11.

Answer:

Question 12.

Answer:
No common tangent because two circles do not intersect at one point.

Question 13.

Answer:

Question 14.

Answer:
One common tangent.

In Exercises 15 – 18, tell whether the common tangent is internal or external.

Question 15.

Answer:

Question 16.

Answer:
The common tangent is the internal tangent because it intersects the segment that joins the centres of two circles.

Question 17.

Answer:

Question 18.

Answer:
The common tangent is the internal tangent because it intersects the segment that joins the centres of two circles.

In Exercises 19 – 22, tell whether \(\overline{A B}\) is tangent to ⊙C. Explain your reasoning.

Question 19.

Answer:

Question 20.

Answer:
Use the converse of the Pythagorean theorem
18² _____________ 15² + 9²
324 _____________ 225 + 81
324 ≠ 304
△ ACB is not a right angled triangle.
So, \(\overline{A B}\) is not tangent to ⊙C at B.

Question 21.

Answer:

Question 22.

Answer:
Use the converse of the Pythagorean theorem
8² _____________ 12² + 16²
64 _____________ 144 + 256
64 ≠ 400
△ ACB is not a right angled triangle.
So, \(\overline{A B}\) is not tangent to ⊙C at B.

In Exercises 23 – 26, point B is a point of tangency. Find the radius r of ⊙C.

Question 23.

Answer:

Question 24.

Answer:
(r + 6)² = r² + 9²
r² + 12r + 36 = r² + 81
12r = 81 – 36
12r = 45
r = \(\frac { 15 }{ 4 } \)
Therefore, the radius of ⊙C is \(\frac { 15 }{ 4 } \)

Question 25.

Answer:

Question 26.

Answer:
(r + 18)² = r² + 30²
r² + 36r + 324 = r² + 900
36r = 900 – 324
36r = 576
r = 16
Therefore, the radius of ⊙C is 16

CONSTRUCTION
In Exercises 27 and 28. construct ⊙C with the given radius and point A outside of ⊙C. Then construct a line tangent to ⊙C that passes through A.

Question 27.
r = 2 in.
Answer:

Question 28.
r = 4.5 cm

Answer:

In Exercises 29 – 32, points B and D are points of tangency. Find the value(s) of x.

Question 29.

Answer:

Question 30.

Answer:
3x + 10 = 7x – 6
7x – 3x = 10 + 6
4x = 16
x = 4

Question 31.

Answer:

Question 32.

Answer:
2x + 5 = 3x² + 2x – 7
3x² = 5 + 7
3x² = 12
x² = 4
x = ±2

Question 33.
ERROR ANALYSIS
Describe and correct the error in determining whether \(\overline{X Y}\) is tangent to ⊙Z.

Answer:

Question 34.
ERROR ANALYSIS
Describe and correct the error in finding the radius of ⊙T.

Answer:
39² = 36² + 15²
So, 15 is the diameter.
The radius is \(\frac { 15 }{ 2 } \).

Question 35.
ABSTRACT REASONING
For a point outside of a circle, how many lines exist tangent to the circle that pass through the point? How many such lines exist for a point on the circle? inside the circle? Explain your reasoning.
Answer:

Question 36.
CRITICAL THINKING
When will two lines tangent to the same circle not intersect? Justify your answer.

Answer:

Using tangent line to circle theorem, it follow that the angle between tangent and radius is a right angle. Let’s draw these tangents at the two ends of the same diameter. We can observe a diameter AD like a transverzal of these tangents.

Question 37.
USING STRUCTURE
Each side of quadrilateral TVWX is tangent to ⊙Y. Find the perimeter of the quadrilateral.

Answer:

Question 38.
LOGIC
In ⊙C, radii \(\overline{C A}\) and \(\overline{C B}\) are perpendicular. are tangent to ⊙C.

a. Sketch ⊙C, \(\overline{C A}\), \(\overline{C B}\), .
Answer:

b. What type of quadrilateral is CADB? Explain your reasoning.
Answer:

Question 39.
MAKING AN ARGUMENT
Two hike paths are tangent to an approximately circular pond. Your class is building a nature trail that begins at the intersection B of the bike paths and runs between the bike paths and over a bridge through the center P of the pond. Your classmate uses the Converse of the Angle Bisector Theorem (Theorem 6.4) to conclude that the trail must bisect the angle formed by the bike paths. Is your classmate correct? Explain your reasoning.

Answer:

Question 40.
MODELING WITH MATHEMATICS
A bicycle chain is pulled tightly so that \(\overline{M N}\) is a common tangent of the gears. Find the distance between the centers of the gears.

Answer:
height h = 4.3 – 1.8
h = 2.5
x² = MN² + h²
x² = 17.6² + 2.5²
x² = 316.01
x = 17.8
Therefore, the distance between the centre of the gear is 17.8 in.

Question 41.
WRITING
Explain why the diameter of a circle is the longest chord of the circle.
Answer:

Question 42.
HOW DO YOU SEE IT?
In the figure, \(\vec{P}\)A is tangent to the dime. \(\vec{P}\)C is tangent to the quarter, and \(\vec{P}\)B is a common internal tangent. How do you know that \(\overline{P A} \cong \overline{P B} \cong \overline{P C}\)

Answer:

Question 43.
PROOF
In the diagram, \(\overline{R S}\) is a common internal tangent to ⊙A and ⊙B. Prove that \(\frac{\Lambda C}{B C}=\frac{R C}{S C}\)

Answer:

Question 44.
THOUGHT PROVOKING
A polygon is circumscribed about a circle when every side of the polygon is tangent to the circle. In the diagram. quadrilateral ABCD is circumscribed about ⊙Q. Is it always true that AB + CD = AD + BC? Justify your answer.

Answer:

Question 45.
MATHEMATICAL CONNECTIONS
Find the values of x and y. Justify your answer.

Answer:

Question 46.
PROVING A THEOREM
Prove the External Tangent Congruence Theorem (Theorem 10.2).

Given \(\overline{S R}\) and \(\overline{S T}\) are tangent to ⊙P.
Prove \(\overline{S R} \cong \overline{S T}\)

Answer:

∠PRS and ∠PTS are the right angles. So the legs of circles are congruent.
Therefore, \(\overline{S R} \cong \overline{S T}\)

Question 47.
PROVING A THEOREM
Use the diagram to prove each part of the biconditional in the Tangent Line to Circle Theorem (Theorem 10.1 ).

a. Prove indirectly that if a line is tangent to a circle, then it is perpendicular to a radius. (Hint: If you assume line m is not perpendicular to \(\overline{Q P}\), then the perpendicular segment from point Q to line m must intersect line m at some other point R.)
Ghen Line m is tangent to ⊙Q at point P.
Prove m ⊥ \(\overline{Q P}\)
b. Prove indirectly that if a line is perpendicular to a radius at its endpoint, then the line is tangent to the circle.
Gien m ⊥ \(\overline{Q P}\)
Prove Line m is tangent to ⊙Q.
Answer:

Question 48.
REASONING
In the diagram, AB = AC = 12, BC = 8, and all three segments are Langent to ⊙P. What is the radius of ⊙P? Justify your answer.

Answer:

Maintaining Mathematical Proficiency

Find the indicated measure.

Question 49.
m∠JKM

Answer:

Question 50.
AB

Answer:
AC = AB + BC
10 = AB + 7
AB = 10 – 7
AB = 3

10.2 Finding Arc Measures

Exploration 1

Measuring Circular Arcs

Work with a partner: Use dynamic geometry software to find the measure of \(\widehat{B C}\). Verify your answers using trigonometry.

a.

Points
A(0, 0)
B(5, 0)
C(4, 3)

Answer:
30 degrees

b.

Points
A(0, 0)
B(5, 0)
C(3, 4)

Answer:
60 degrees

c.

Points
A(0, 0)
B(4, 3)
C(3, 4)

Answer:
15 degrees

d.

Points
A(0, 0)
B(4, 3)
C(- 4, 3)

Answer:
90 degrees

Communicate Your Answer

Question 2.
How are circular arcs measured?
Answer:

Question 3.
Use dynamic geometry software to draw a circular arc with the given measure.
USING TOOLS STRATEGICALLY
To be proficient in math, you need to use technological tools to explore and deepen your understanding of concepts.
a. 30°

Answer:

b. 45°

Answer:

c. 60°
Answer:

d. 90°

Answer:

Lesson 10.2 Finding Arc Measures

Monitoring Progress

Identify the given arc as a major arc, minor arc, or semicircle. Then find the measure of the arc.

Question 1.
\(\widehat{T Q}\)

Answer:
\(\widehat{T Q}\) is a minor arc.
\(\widehat{T Q}\) = 120°

Question 2.
\(\widehat{Q R T}\)

Answer:
\(\widehat{Q R T}\)

Question 3.
\(\widehat{T Q R}\) is a major arc.
\(\widehat{Q R T}\) = QR + RS + ST
RS = 360° – (60 + 120 + 80)
= 360 – 260 = 100°
So, \(\widehat{Q R T}\) = 60° + 100° + 80°
\(\widehat{Q R T}\) = 240°

Answer:

Question 4.
\(\widehat{Q S}\)

Answer:
\(\widehat{Q S}\) = QR + RS
= 60 + 100 = 160°
Therefore, \(\widehat{Q S}\) = 160° and it is a minor arc.

Question 5.
\(\widehat{T S}\)

Answer:
\(\widehat{T S}\) = 80° and it is a minor arc.

Question 6.
\(\widehat{R S T}\)

Answer:
\(\widehat{R S T}\) = 100 + 80 = 180
Therefore, \(\widehat{R S T}\) = 180° and it is a minor arc.

Tell whether the red arcs are congruent. Explain why or why not.

Question 7.

Answer:
\(\widehat{A B}\), \(\widehat{C D}\) are congruent as they measure same radius and same arc length.

Question 8.

Answer:
\(\widehat{M N}\), \(\widehat{P Q}\) are not congruent as they measure different radius.

Exercise 10.2 Finding Arc Measures

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Copy and complele:
If ∠ACB and ∠DCE are congruent central angles of ⊙C, then \(\widehat{A B}\) and \(\widehat{D E}\) arc.
Answer:

Question 2.
WHICH ONE DOESNT BELONG?
Which circle does not belong with the other three? Explain your reasoning.

Answer:
We know that 1 ft = 12 in
So, the fourth circle does not belong to the other three as its diameter is different.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, name the red minor arc and find its measure. Then name the blue major arc and find its measure.

Question 3.

Answer:

Question 4.

Answer:
The minor arc \(\widehat{E F}\) = 68° and major arc \(\widehat{F G E}\) = 360° – 68° = 292°.

Question 5.

Answer:

Question 6.

Answer:
The minor arc is \(\widehat{M N}\) = 170°, major arc \(\widehat{N P M}\) = 360° – 170° = 190°.

In Exercises 7 – 14. identify the given arc as a major arc, minor arc, or semicircle. Then find the measure of the arc.

Question 7.
\(\widehat{B C}\)
Answer:

Question 8.
\(\widehat{D C}\)

Answer:
\(\widehat{D C}\) is a minor arc and it has a measure of 65°.

Question 9.
\(\widehat{E D}\)
Answer:

Question 10.
\(\widehat{A E}\)

Answer:
\(\widehat{A E}\) is a minor arc and it has a measure of 70°.

Question 11.
\(\widehat{E A B}\)
Answer:

Question 12.
\(\widehat{A B C}\)

Answer:
\(\widehat{A B C}\) is a semicircle and it has a measure of 180°.

Question 13.
\(\widehat{B A C}\)
Answer:

Question 14.
\(\widehat{E B D}\)

Answer:
\(\widehat{E B D}\) is a major arc and it has a measure of 315°.

In Exercises 15 and 16, find the measure of each arc.

Question 15.

a. \(\widehat{J L}\)

b. \(\widehat{K M}\)

c. \(\widehat{J L M}\)

d. \(\widehat{J M}\)
Answer:

Question 16.

a. \(\widehat{R S}\)

Answer:
\(\widehat{R S}\) = \(\widehat{Q R S}\) – \(\widehat{Q R}\)
= 180 – 42
= 138°
So, \(\widehat{R S}\) = 138°

b. \(\widehat{Q R S}\)

Answer:
\(\widehat{Q R S}\) = 180°

c. \(\widehat{Q S T}\)

Answer:
\(\widehat{Q S T}\) = \(\widehat{Q R S}\) + \(\widehat{S T}\)
= 180 + 42 = 222
So, \(\widehat{Q S T}\) = 222°

d. \(\widehat{Q T}\)

Answer:
\(\widehat{Q T}\) = 360 – (42 + 138 + 42)
= 360 – (222)
= 138°
\(\widehat{Q T}\) = 138°

Question 17.
MODELING WITH MATHEMATICS
A recent survey asked high school students their favorite type of music. The results are shown in the circle graph. Find each indicated arc measure.

a. m\(\widehat{A E}\)
b. m\(\widehat{A C E}\)
c. m\(\widehat{G D C}\)
d. m\(\widehat{B H C}\)
e. m\(\widehat{F D}\)
f. m\(\widehat{F B D}\)
Answer:

Question 18.
ABSTRACT REASONING
The circle graph shows the percentages of students enrolled in fall Sports at a high school. Is it possible to find the measure of each minor arc? If so, find the measure 0f the arc for each category shown. If not, explain why it is not possible.

Answer:
Soccer angle = 30% of 360 = 108°
Volleyball angle = 15% of 360 = 54°
Cross-country angle = 20% of 360 = 72°
None angle = 15% of 360 = 54°
Football angle = 20% of 360 = 72°

In Exercises 19 – 22, tell whether the red arcs are congruent. Explain why or why not.

Question 19.

Answer:

Question 20.

Answer:
\(\widehat{L P}\) and \(\widehat{M N}\) are not congruet because they are not in the same circle.

Question 21.

Answer:

Question 22.

Answer:
\(\widehat{R S Q}\), \(\widehat{F G H}\) are not congruent because those two circles have different radii.

MATHEMATICAL CONNECTIONS
In Exercises 23 and 24. find the value of x. Then find the measure of the red arc.

Question 23.

Answer:

Question 24.

Answer:
4x + 6x + 7x + 7x = 360
24x = 360
x = 15
m\(\widehat{R S T}\) = 6(15) + 7(15)
= 90 + 105 = 195°
So, m\(\widehat{R S T}\) = 195°

Question 25.
MAKING AN ARGUMENT
Your friend claims that any two arcs with the same measure are similar. Your cousin claims that an two arcs with the same measure are congruent. Who is correct? Explain.
Answer:

Question 26.
MAKING AN ARGUMENT
Your friend claims that there is not enough information given to find the value of x. Is your friend correct? Explain your reasoning.

Answer:
My friend is wrong.
4x + x + x + 4x = 360
10x = 360
x = 36°

Question 27.
ERROR ANALYSIS
Describe and correct the error in naming the red arc.

Answer:

Question 28.
ERROR ANALYSIS
Describe and correct the error in naming congruent arc.

Answer:
\(\widehat{J K}\), \(\widehat{N P}\) are not congruent because those two arcs are from different circles.

Question 29.
ATTENDING TO PRECISION
Two diameters of ⊙P are \(\widehat{A B}\) and \(\widehat{C D}\). Find m\(\widehat{A C D}\) and m\(\widehat{A C}\) when m\(\widehat{A D}\) = 20°.
Answer:

Question 30.
REASONING
In ⊙R, m\(\widehat{A B}\) = 60°, m\(\widehat{B C}\) = 25°. m\(\widehat{C D}\) = 70°, and m\(\widehat{D E}\) = 20°. Find two possible measures of \(\widehat{A E}\).

Answer:
\(\widehat{A E}\) = 360 – (\(\widehat{A B}\) + \(\widehat{B C}\) + \(\widehat{C D}\) + \(\widehat{D E}\))
= 360 – (60 + 25 + 70 + 20)
= 360 – (175)
= 185
\(\widehat{A E}\) = \(\widehat{A B}\) + \(\widehat{B C}\) + \(\widehat{C D}\) + \(\widehat{D E}\)
= 60 + 25 + 70 + 20 = 175
So, the two possibilities of \(\widehat{A E}\) are 185°, 175°

Question 31.
MODELING WITH MATHEMATICS
On a regulation dartboard, the outermost circle is divided into twenty congruent sections. What is the measure of each arc in this circle?

Answer:

Question 32.
MODELING WITH MATHEMATICS
You can use the time zone wheel to find the time in different locations across the world. For example, to find the time in Tokyo when it is 4 P.M. in San Francisco, rotate the small wheel until 4 P.M. and San Francisco line up, as shown. Then look at Tokyo to see that it is 9 A.M. there.

a. What is the arc measure between each time zone 0n the wheel?

Answer:
As the circle is divided into 24 sectors, each time zone angle = \(\frac { 360 }{ 24 } \) = 15°

b. What is the measure of the minor arc from the Tokyo zone to the Anchorage zone?

Answer:
The measure of the minor arc from the Tokyo zone to the Anchorage zone = 15 + 15 + 15 + 15 + 15 + 15
= 90°

c. If two locations differ by 180° on the wheel, then it is 3 P.M. at one location when it is _________ at the other location.

Answer:
Kuwaiti city.

Question 33.
PROVING A THEOREM
Write a coordinate proof of the Similar Circles Theorem (Theorem 10.5).
Given ⊙O with center O(0, 0) and radius r.
⊙A with center A(a, 0) and radius s
Prove ⊙O ~ ⊙A

Answer:

Question 34.
ABSTRACT REASONING
Is there enough information to tell whether ⊙C ≅ ⊙D? Explain your reasoning.

Answer:
Both circles ⊙C and ⊙D have the same radius so those circles are congruent.

Question 35.
PROVING A THEOREM
Use the diagram to prove each part of the biconditional in the Congruent Circles Theorem (Theorem 10.3).

a. Given \(\overline{A C} \cong \overline{B D}\)
Prove ⊙A ≅ ⊙B
b. Given ⊙A ≅ ⊙B
prove \(\overline{A C} \cong \overline{B D}\)
Answer:

Question 36.
HOW DO YOU SEE IT?
Are the circles on the target similar or congruent? Explain your reasoning.

Answer:

Question 37.
PROVING A THEOREM
Use the diagram to prove each part of the biconditional in the Congruent Central Angles Theorem (Theorem 10.4).

a. Given ∠ABC ≅ ∠DAE
Prove \(\widehat{B C}\) ≅ \(\widehat{D E}\)
b. Given \(\widehat{B C}\) ≅ \(\widehat{D E}\)
Prove ∠ABC ≅ ∠DAE
Answer:

Question 38.
THOUGHT PROVOKING
Write a formula for the length of a circular arc. Justify your answer.

Answer:
The formula to find the length of a circular arc is radius x angle.

Maintaining Mathematical Proficiency

Find the value of x. Tell whether the side lengths form a Pythagorean triple.

Question 39.

Answer:

Question 40.

Answer:
x² = 13² + 13²
= 169 + 169
= 338
x = 13√2

Question 41

Answer:

Question 42.

Answer:
14² = x² + 10²
196 = x² + 100
x² = 196 – 100
x² = 96
x = 4√6

10.3 Using Chords

Exploration 1

Drawing Diameters

Work with a partner: Use dynamic geometry software to construct a circle of radius 5 with center at the origin. Draw a diameter that has the given point as an endpoint. Explain how you know that the chord you drew is a diameter.
a. (4, 3)
b. (0, 5)
c. (-3, 4)
d. (-5, 0)

Answer:

Exploration 2

Writing a Conjecture about Chords

Work with a partner. Use dynamic geometry software to construct a chord \(\overline{B C}\) of a circle A. Construct a chord on the perpendicular bisector of \(\overline{B C}\). What do you notice? Change the original chord and the circle several times. Are your results always the same? Use your results to write a conjecture.
LOOKING FOR STRUCTURE
To be proficient in math, you need to look closely to discern a pattern or structure.

Answer:

Exploration 3

A Chord Perpendicular to a Diameter

Work with a partner. Use dynamic geometry software to construct a diameter \(\overline{B C}\) of a circle A. Then construct a chord \(\overline{D E}\) perpendicular to \(\overline{B C}\) at point F. Find the lengths DF and EF. What do you notice? Change the chord perpendicular to \(\overline{B C}\) and the circle several times. Do you always get the same results? Write a conjecture about a chord that is perpendicular to a diameter of a circle.

Answer:

Communicate Your Answer

Question 4.
What are two ways to determine when a chord is a diameter of a circle?

Answer:
If a chord passes through the centre of the circle, then it is the diameter of a circle.
The longest chord of the circle is the diameter of a circle.

Lesson 10.3 Using Chords

Monitoring Progress

In Exercises 1 and 2, use the diagram of ⊙D.

Question 1.
If m\(\widehat{A B}\) = 110°. find m\(\widehat{B C}\).

Answer:
Because AB and BC are congruent chords in congruent circles, the corresponding minor arcs \(\widehat{A B}\), \(\widehat{B C}\) are congruent by the congruent corresponding chords theorem.
So, \(\widehat{A B}\) = \(\widehat{B C}\)
\(\widehat{B C}\) = 110°

Question 2.
If m\(\widehat{A C}\) = 150° find m\(\widehat{A B}\).

Answer:
\(\widehat{A C}\) = 360 – (\(\widehat{A B}\) + \(\widehat{B C}\))
150 = 360 – 2(\(\widehat{A B}\))
2(\(\widehat{A B}\)) = 360 – 150 = 210
\(\widehat{A B}\) = 105°

In Exercises 3 and 4. find the indicated length or arc measure.

Question 3.
CE

Answer:
CE = 5 + 5
= 10 units

Question 4.
m\(\widehat{C E}\)

Answer:
m\(\widehat{C E}\) = 9x + 180 – x = 180 – 8x
m\(\widehat{C E}\) = 180 – 8x

Question 5.
In the diagram, JK = LM = 24, NP = 3x, and NQ = 7x – 12. Find the radius of ⊙N

Answer:

Exercise 10.3 Using Chords

Vocabulary and Core Concept Check

Question 1.
WRITING
Describe what it means to bisect a chord.
Answer:

Question 2.
WRITING
Two chords of a circle are perpendicular and congruent. Does one of them have to be a diameter? Explain your reasoning.

Answer:

Imagine a line segment of length 3 units, AB.

A second congruent segment of length 3 that is perpendicular to AB called CD.

Circumscribe both these line segments and note that AB and CD are now chords.
While both chords are perpendicular and congruent, neither chord is a diameter. Thus, it is possible to have two chords of this type with neither one diameter of the circle.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, find the measure of the red arc or chord in ⊙C.

Question 3.

Answer:

Question 4.

Answer:
Arc length = radius x angle
= 5 x 34 = 170

Question 5.

Answer:

Question 6.

Answer:
As the two circles radius is the same and the angle is the same so the chord length is 11 units.

In Exercise 7-10, find the value of x.

Question 7.

Answer:

Question 8.

Answer:
By the perpendicular bisector theorem RS = ST
x = 40°

Question 9.

Answer:

Question 10.

Answer:
5x + 2 = 7x – 12
7x – 5x = 2 + 12
2x = 14
x = 7

Question 11.
ERROR ANALYSIS
Describe and correct the error in reasoning.

Answer:

Question 12.
PROBLEM SOLVING
In the cross section of the submarine shown, the control panels are parallel and the same length. Describe a method you can use to find the center of the cross section. Justify your method.

Answer:

In Exercises 13 and 14, determine whether \(\overline{A B}\) is a diameter of the circle. Explain your reasoning.

Question 13.

Answer:

Question 14.

Answer:
5² = 3² + x²
25 = 9 + x²
x² = 25 – 9
x = 4
So, AB is not diameter of the circle.

In Exercises 15 and 16, find the radius of ⊙Q.

Question 15.

Answer:

Question 16.

Answer:
AD = BC
4x + 4 = 6x – 6
6x – 4x = 4 + 6
2x = 10
x = 5
BC = 6(5) – 6 = 30 – 6 = 24
QC² = 5² + 12²
= 25 + 144 = 169
QC = 13
Therefore, the radius is 13.

Question 17.
PROBLEM SOLVING
An archaeologist finds part of a circular plate. What was the diameter of the plate to the nearest tenth of an inch? Justify your answer.

Answer:

Question 18.
HOW DO YOU SEE IT?
What can you conclude from each diagram? Name a theorem that justifies your answer.
a.

Answer:
Perpendicular chord bisector converse theorem.

b.

Answer:
Congruent Corresponding Chords theorem

c.

Answer:
Perpendicular chord bisector theorem

d.

Answer:
Equidistant chords theorem

Question 19.
PROVING A THEOREM
Use the diagram to prove each part of the biconditional in the Congruent Corresponding Chords Theorem (Theorem 10.6).

a. Given \(\overline{A B}\) and \(\overline{C D}\) are congruent chords.
Prove \(\widehat{A B} \cong \widehat{C D}\)
b. Given \(\widehat{A B} \cong \widehat{C D}\)
Prove \(\overline{A B}\) ≅ \(\overline{C D}\)
Answer:

Question 20.
MATHEMATICAL CONNECTIONS
In ⊙P, all the arcs shown have integer measures. Show that x must be even.

Answer:

Question 21.
REASONING
In ⊙P. the lengths of the parallel chords are 20, 16, and 12. Find m\(\widehat{A B}\). Explain your reasoning.

Answer:

Question 22.
PROVING A THEOREM
Use congruent triangles to prove the Perpendicular Chord Bisector Theorem (Theorem 10.7).

Given \(\overline{E G}\) is a diameter of ⊙L.
\(\overline{E G}\) ⊥ \(\overline{D F}\)
Prove \(\overline{D C}\) ≅ \(\overline{F C}\), \(\widehat{D G} \cong \widehat{F G}\)

Answer:
Let L be the centre of the circle
draw any chord DF on the circle
As DF passes through LG.
The length of DC is the same as FC.

Question 23.
PROVING A THEOREM
Write a proof of the Perpendicular Chord Bisector Converse (Theorem 10.8).

Given \(\overline{Q S}\) is a perpendicular bisector of \(\overline{R T}\).
Prove \(\overline{Q S}\) is a diameter of the circle L.
(Hint: Plot the center L and draw △LPT and △LPR.)
Answer:

Question 24.
THOUGHT PROVOKING
Consider two chords that intersect at point P. Do you think that \(\frac{A P}{B P}=\frac{C P}{D P}\)? Justify your answer.

Answer:

Question 25.
PROVING A THEOREM
Use the diagram with the Equidistant Chords Theorem (Theorem 10.9) to prove both parts of the biconditional of this theorem.

\(\overline{A B}\) ≅ \(\overline{C D}\) if and only if EF = EG
Answer:

Question 26.
MAKING AN ARGUMENT
A car is designed so that the rear wheel is only partially visible below the body of the car. The bottom edge of the panel is parallel [o the ground. Your friend claims that the point where the tire touches the ground bisects \(\widehat{A B}\). Is your friend correct? Explain your reasoning.

Answer:

Maintaining Mathematical Proficiency

Find the missing interior angle measure.

Question 27.
Quadrilateral JKLW has angle measures m∠J = 32°, m∠K = 25°, and m∠L = 44°. Find m∠M.
Answer:

Question 28.
Pentagon PQRST has angle measures m∠P = 85°, m∠Q = 134°, m∠R = 97°, and m∠S =102°.
Find m∠T.

Answer:
The sum of interior angles of a pentagon = 540°
m∠T = 540 – (85 + 134 + 97 + 102)
= 540 – 418 = 122
m∠T = 122°.

10.1 – 10.3 Quiz

In Exercises 1 – 6, use the diagram. (Section 10.1)

Question 1.
Name the circle.

Answer:
The circle has a chord, diameter and tangent.

Question 2.
Name a radius.

Answer:
NP is the radius of the circle.

Question 3.
Name a diameter.

Answer:
KN is the diameter of the circle.

Question 4.
Name a chord.

Answer:
JL is the chord

Question 5.
Name a secant.

Answer:
SN is the secant

Question 6.
Name a tangent.

Answer:
QR is the tangent.

Find the value of x.

Question 7.

Answer:
(9 + x)² = x² + 15²
81 + 18x + x² = x² + 225
18x = 225 – 81
18x = 144
x = 8

Question 8.

Answer:
6x – 3 = 3x + 18
6x – 3x = 18 + 3
3x = 21
x = 7

Identify the given arc as a major arc, minor arc, or semicircle. Then find the measure of the arc.

Question 9.
\(\widehat{A E}\)

Answer:
\(\widehat{A E}\) = 180 – 36
= 144
So, \(\widehat{A E}\) = 144°

Question 10.
\(\widehat{B C}\)

Answer:
\(\widehat{B C}\) = 180 – (67 + 70)
= 180 – 137 = 43
So, \(\widehat{B C}\) = 43°

Question 11.
\(\widehat{A C}\)

Answer:
\(\widehat{A C}\) = 43 + 67 = 110°

Question 12.
\(\widehat{A C D}\)

Answer:
\(\widehat{A C D}\) = 180°

Question 13.
\(\widehat{A C E}\)

Answer:
\(\widehat{A C E}\) = 180 + 36 = 216°

Question 14.
\(\widehat{B E C}\)

Answer:
\(\widehat{B E C}\) = 70 + 36 + 43 = 149°

Tell whether the red arcs are congruent. Explain why or why not.

Question 15.

Answer:
As two chords pass through the centre of the circle. Those two red arcs are congruent.

Question 16.

Answer:
Red arcs are not congruent because the radius of the two circles is different.

Question 17.
Find the measure of the red arc in ⊙Q.

Answer:

Question 18.
In the diagram. AC = FD = 30, PG = x + 5, and PJ = 3x – 1. Find the radius of ⊙P.

Answer:

Question 19.
A circular clock can be divided into 12 congruent sections.

a. Find the measure of each arc in this circle.
Answer:
The measure of each arc = \(\frac { 360 }{ 12 } \) = 30°

b. Find the measure of the minor arc formed by the hour and minute hands when the times is 7:00.
Answer:
When the time is 7:00 the minute hand is at 12 and hour hand is at 7 and so the minor arc is subtended by 12 – 7 = 5 of these sections and so the angle subtended is 30 x 5 = 150°

c. Find a time at which the hour and minute hands form an arc that is congruent to the arc in part (b).
Answer:
A minor arc is equal to 150° can be formed by multiple placements of the hour and the minute hand. One of them can be the time 5:00 when the minute hand is at 12 and the hour hand is at 5.

10.4 Inscribed Angles and Polygons

Exploration 1

Inscribed Angles and Central Angles

work with a partner: Use dynamic geometry software.

Sample

a. Construct an inscribed angle in a circle. Then construct the corresponding central angle.
Answer:

b. Measure both angles. How is the inscribed angle related to its intercepted arc?
Answer:

c. Repeat parts (a) and (b) several times. Record your results in a table. Write a conjecture about how an inscribed angle is related to its intercepted arc.
ATTENDING TO PRECISION
To be proficient in math, you need to communicate precisely with others.
Answer:

Exploration 2

A Quadrilateral with Inscribed Angles

work with a partner: Use dynamic geometry software.

Sample

a. Construct a quadrilateral with each vertex on a circle.
Answer:

b. Measure all four angles. What relationships do you notice?
Answer:

c. Repeat parts (a) and (b) several times. Record your results in a table. Then write a conjecture that summarizes the data.
Answer:

Communicate Your Answer

Question 3.
How are inscribed angles related to their intercepted arcs? How are the angles of an inscribed quadrilateral related to each other?
Answer:

Question 4.
Quadrilateral EFGH is inscribed in ⊙C. and m ∠ E = 80°. What is m ∠ G? Explain.

Answer:
m ∠ E + m ∠ H = 80 + 80 = 160°
m ∠ E + m ∠ H + m ∠ G + m ∠ F = 360
160° + m ∠ G + m ∠ F = 360
m ∠ G + m ∠ F = 360 – 160 = 200
m ∠ G = 100°

Lesson 10.4 Inscribed Angles and Polygons

Monitoring Progress

Find the measure of the red arc or angle.

Question 1.

Answer:
m∠G = \(\frac { 90 }{ 2 } \) = 45°

Question 2.

Answer:
\(\widehat{T V}\) = 2 • 38 = 76°

Question 3.

Answer:
m∠W = 72°

Find the value of each variable.

Question 4.

Answer:
x° = 90°
y° = 180 – (40 + 90) = 180 – 130
y° = 50°

Question 5.

Answer:
∠B + ∠D = 180
∠B + 82 = 180
x° = 98°
∠C + ∠A = 180
68 + y° = 180
y° = 112°

Question 6.

Answer:
∠S + ∠U = 180°
c + 2c – 6 = 180
3c = 186
c = 62°
∠T + ∠V = 180°
10x + 8x = 180
18x = 180
x = 10°

Question 7.
In Example 5, explain how to find locations where the left side of the statue is all that appears in your camera’s field of vision.
Answer:

Exercise 10.4 Inscribed Angles and Polygons

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
If a circle is circumscribed about a polygon, then the polygon is an ___________ .
Answer:

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different?
Find “both” answers.

Find m∠ABC.
Answer:
m∠ABC = 60°

Find m∠AGC.
Answer:
m∠AGC = 180 – (25 + 25)
= 180 – 50 = 130°

Find m∠AEC.
Answer:
m∠AEC = 180 – (50 + 50)
= 180 – 100 = 80°

Find m∠ADC.
Answer:
m∠ADC = 180 – (25 + 50)
= 180 – 75 = 105°

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 8, find the indicated measure.

Question 3.
m∠A

Answer:

Question 4.
m∠G

Answer:
m∠G = 360 – (70 + 120)
= 360 – 190 = 170°

Question 5.
m ∠ N

Answer:

Question 6.
m\(\widehat{R S}\)

Answer:
m\(\widehat{R S}\) = 2 • 67 = 134°

Question 7.
m\(\widehat{V U}\)

Answer:

Question 8.
m\(\widehat{W X}\)

Answer:
m\(\widehat{W X}\) = \(\frac { 75 }{ 2 } \) = 37.5

In Exercises 9 and 10, name two pairs of congruent angles.

Question 9.

Answer:

Question 10.

Answer:
m∠W = m∠Z, m∠X = m∠Y

In Exercises 11 and 12, find the measure of the red arc or angle.

Question 11.

Answer:

Question 12.

Answer:
\(\widehat{P S}\) = 2 • 40 = 80

In Exercises 13 – 16, find the value of each variable.

Question 13.

Answer:

Question 14.

Answer:
m∠E + m∠G = 180
m + 60 = 180
m = 120°
m∠D + m∠F = 180
60 + 2k = 180
k = 60°

Question 15.

Answer:

Question 16.

Answer:
3x° = 90°
x° = 30°
2y° + 90° + 34° = 180°
2y° + 124° = 180°
2y° = 56°
y° = 28°

Question 17.
ERROR ANALYSIS
Describe and correct the error in finding m\(\widehat{B C}\).

Answer:

Question 18.
MODELING WITH MATHEMATICS
A carpenter’s square is an L-shaped tool used to draw right angles. You need to cut a circular piece of wood into two semicircles. How can you use the carpenter’s square to draw a diameter on the circular piece of wood?

Answer:
Recall that when a right triangle is inscribed in a circle, then the hypotenuse is the diameter of the circle. Simply use the carpenter’s square to inscribe it into the circle. The hypotenuse formed by both legs of the square should provide a diameter for the circle.

MATHEMATICAL CONNECTIONS
In Exercises 19 – 21, find the values of x and y. Then find the measures of the interior angles of the polygon.

Question 19.

Answer:

Question 20.

Answer:
∠B + ∠C = 180
14x + 4x = 180
18x = 180°
x = 10°
∠A + ∠D = 180
9y + 24y = 180
33y = 180°
y = 5.45°
∠A = 130.9°, ∠B = 40°, ∠C = 140°, ∠D = 49°

Question 21.

Answer:

Question 22.
MAKING AN ARGUMENT
Your friend claims that ∠PTQ ≅ ∠PSQ ≅ ∠PRQ. Is our friend correct? Explain your reasoning.

Answer:
Yes, my friend is correct.
∠PTQ ≅ ∠PSQ ≅ ∠PRQ is correct according to the inscribed angles of a circle theorem.

Question 23.
CONSTRUCTION
Construct an equilateral triangle inscribed in a circle.
Answer:

Question 24.
CONSTRUTION
The side length of an inscribed regular hexagon is equal to the radius of the circumscribed circle. Use this fact to construct a regular hexagon inscribed in a circle.

Answer:
As the side length is equal to the radius. Draw a line representing the radius and draw a chord different chords in the form of hexagons of the radius of the circle.

REASONING
In Exercises 25 – 30, determine whether a quadrilateral of the given type can always be inscribed inside a circle. Explain your reasoning.

Question 25.
Square
Answer:

Question 26.
rectangle
Answer:
yes, angles are right angles.

Question 27.
parallelogram
Answer:

Question 28.
kite
Answer:
No.

Question 29.
rhombus
Answer:

Question 30.
isosceles trapezoid
Answer:
Yes, the opposite angles are always supplementary.

Question 31.
MODELING WITH MATHEMATICS
Three moons, A, B, and C, are in the same circular orbit 1,00,000 kilometers above the surface of a planet. The planet is 20,000 kilometers in diameter and m∠ABC = 90°. Draw a diagram of the situation. How far is moon A from moon C?
Answer:

Question 32.
MODELING WITH MATHEMATICS
At the movie theater. you want to choose a seat that has the best viewing angle, so that you can be close to the screen and still see the whole screen without moving your eyes. You previously decided that seat F7 has the best viewing angle, but this time someone else is already sitting there. Where else can you sit so that your seat has the same viewing angle as seat F7? Explain.

Answer:

Question 33.
WRITING
A right triangle is inscribed in a circle, and the radius of the circle is given. Explain how to find the length of the hypotenuse.
Answer:

Question 34.
HOW DO YOU SEE IT?
Let point Y represent your location on the soccer field below. What type of angle is ∠AYB if you stand anywhere on the circle except at point A or point B?

Answer:

Question 35.
WRITING
Explain why the diagonals of a rectangle inscribed in a circle are diameters of the circle.
Answer:

Question 36.
THOUGHT PROVOKING
The figure shows a circle that is circumscribed about ∆ABC. Is it possible to circumscribe a circle about any triangle? Justify your answer.

Answer:
Yes.

Question 37.
PROVING A THEOREM
If an angle is inscribed in ⊙Q. the center Q can be on a side of the inscribed angle, inside the inscribed angle, or outside the inscribed angle. Prove each case of the Measure of an Inscribed Angle Theorem (Theorem 10. 10).

a. Case 1
Given ∠ABC is inscribed in ⊙Q
Let m∠B = x°
Center Q lies on \(\overline{B C}\).
Prove m∠ABC = \(\frac{1}{2}\)m\(\widehat{A C}\)
(Hint: Show that ∆AQB is isosceles. Then write m\(\widehat{A C}\) in terms of x.)

b. Case 2
Use the diagram and auxiliary line to write Given and Prove statements for Case 2. Then write a proof

c. Case 3
Use the diagram and auxiliary line to write Given and Prove statements for Case 3. Then write a proof.

Answer:

Question 38.
PROVING A THEOREM
Write a paragraph proof of the Inscribed Angles of a Circle Theorem (Theorem 10.11). First, draw a diagram and write Given and Prove statements.
Answer:
If two inscribed angles of a circle intercept the same arc, then the angles are congruent.

Question 39.
PROVING A THEOREM
The Inscribed Right Triangle Theorem (Theorem 10.12) is written as a conditional statement and its converse. Write a plan for proof for each statement.
Answer:

Question 40.
PROVING A THEOREM
Copy and complete the paragraph proof for one part of the Inscribed Quadrilateral Theorem (Theorem 10. 13).
Given ⊙C with inscribed quadrilateral DEFG
Prove m ∠ D + m ∠ F = 180°,
m ∠ E + m ∠ G = 180°

By the Arc Addition Postulate (Postulate 10. 1),
m\(\widehat{E F G}\) + ________ = 360° and m\(\widehat{F G D}\) + m\(\widehat{D E F}\) = 360°.
Using the ___________ Theorem. m\(\widehat{E D G}\) = 2m ∠ F, m\(\widehat{E F G}\) = 2m∠D, m\(\widehat{D E F}\) = 2m∠G, and m\(\widehat{F G D}\) = 2m ∠ E. By the Substitution Property of Equality, 2m∠D + ________ = 360°, So _________ . Similarly, __________ .
Answer:
m\(\widehat{E F G}\) + m\(\widehat{E D F}\) = 360° and m\(\widehat{F G D}\) + m\(\widehat{D E F}\) = 360°.
Using the the measure of an inscribed angle Theorem. m\(\widehat{E D G}\) = 2m ∠ F, m\(\widehat{E F G}\) = 2m∠D, m\(\widehat{D E F}\) = 2m∠G, and m\(\widehat{F G D}\) = 2m ∠ E. By the Substitution Property of Equality, 2m∠D + 2m∠G = 360°.

Question 41.
CRITICAL THINKING
In the diagram, ∠C is a right angle. If you draw the smallest possible circle through C tangent to \(\overline{A B}\), the circle will intersect \(\overline{A C}\) at J and \(\overline{B C}\) at K. Find the exact length of \(\overline{J K}\).

Answer:

Question 42.
CRITICAL THINKING
You are making a circular cutting board. To begin, you glue eight 1-inch boards together, as shown. Then you draw and cut a circle with an 8-inch diameter from the boards.

a. \(\overline{F H}\) is a diameter of the circular cutting board. Write a proportion relating GJ and JH. State a theorem in to justify your answer.
Answer:
Each board is 1 inch and FJ spans 6 boards.
\(\overline{F H}\) = 6 inches

b. Find FJ, JH, and GJ. What is the length of the cutting board seam labeled \(\overline{G K}\)?
Answer:
Each board is 1 inch and JH spans 2 boards.
JH = 2 inches
Equation is \(\frac { JH }{ GJ } \) = \(\frac { GJ }{ FJ } \)
\(\frac { 2 }{ GJ } \) = \(\frac { GJ }{ 6 } \)
12 = GJ²
GJ = 2√3
GK = 2(GJ)
GK = 4√3
So, FJ = 6, JH = 2, JG = 2√3, GK = 4√3

Maintaining Mathematical Proficiency

Solve the equation. Check your solution.

Question 43.
3x = 145
Answer:

Question 44.
\(\frac{1}{2}\)x = 63
Answer:
x = 63 • 2
x = 126

Question 45.
240 = 2x
Answer:

Question 46.
75 = \(\frac{1}{2}\)(x – 30)
Answer:
75 • 2 = x – 30
150 + 30 = x
x = 180

10.5 Angle Relationships in Circles

Exploration 1

Angles Formed by a Chord and Tangent Line

Work with a partner: Use dynamic geometry software.

Sample

a. Construct a chord in a circle. At one of the endpoints of the chord. construct a tangent line to the circle.
Answer:

b. Find the measures of the two angles formed by the chord and the tangent line.
Answer:

c. Find the measures of the two circular arcs determined by the chord.
Answer:

d. Repeat parts (a) – (c) several times. Record your results in a table. Then write a conjecture that summarizes the data.
Answer:

Exploration 2

Angles Formed by Intersecting Chords

Work with a partner: Use dynamic geometry software.

sample

a. Construct two chords that intersect inside a circle.
Answer:

b. Find the measure of one of the angles formed by the intersecting chords.
Answer:

c. Find the measures of the arcs intercepted h the angle in part (b) and its vertical angle. What do you observe?
Answer:

d. Repeat parts (a) – (c) several times. Record your results in a table. Then write a conjecture that summarizes the data.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to understand and use stated assumptions, definitions, and previously established results.
Answer:

Communicate Your Answer

Question 3.
When a chord intersects a tangent line or another chord, what relationships exist among the angles and arcs formed?
Answer:

Question 4.
Line m is tangent to the circle in the figure at the left. Find the measure of ∠1.

Answer:
m∠1 = \(\frac { 1 }{ 2 } \) • 148
m∠1 = 74°

Question 5.
Two chords intersect inside a circle to form a pair of vertical angles with measures of 55°. Find the sum of the measures of the arcs intercepted by the two angles.
Answer:
The sum of the measures of the arcs intercepted by the two angles = \(\frac { 1 }{ 2 } \) • 55
= 27.5

Lesson 10.5 Angle Relationships in Circles

Monitoring Progress

Line m is tangent to the circle. Find the indicated measure.

Question 1.
m ∠ 1

Answer:
m ∠ 1 = \(\frac { 1 }{ 2 } \) • 210
m ∠ 1 = 105°

Question 2.
m\(\widehat{R S T}\)

Answer:
m\(\widehat{R S T}\) = 2 • 98 = 196°
m\(\widehat{R S T}\) = 196°

Question 3.
m\(\widehat{X Y}\)

Answer:
m\(\widehat{X Y}\) = \(\frac { 1 }{ 2 } \) • 80
m\(\widehat{X Y}\) = 40°

Find the value of the variable.

Question 4.

Answer:
y° = \(\frac { 1 }{ 2 } \) • (102 + 95)
= 98.5°

Question 5.

Answer:
a° = 2 • 30° + 44°
= 60° + 44° = 104°
So, a° = 104°.

Find the value of x.

Question 6.

Answer:
x° = 180° – 120°
x° = 60°

Question 7.

Answer:
50° = 180° – x°
x° = 180° – 50°
x° = 130°

Question 8.
You are on top of Mount Rainier on a clear day. You are about 2.73 miles above sea level at point B. Find m\(\widehat{C D}\), which represents the part of Earth that you can see.

Answer:
CB and BD are tangents, CB is perpendicular to AB and CD is perpendicular to AD by the tangent line to circle theorem.
△ABC is similar to △ABD by the hypotenuese leg congruence theorem.
∠CBA is similar to ∠ABD. So, m∠CBA = 74.5°, m∠CBD = 2 • 74.5° = 149°
m∠CBD = 180° – m∠CAD
m∠CBD = 180° – CD
149° = 180° – CD
CD = 31°
The part of earth you can see

Exercise 10.5 Angle Relationships in Circles

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
Points A, B, C, and D are on a circle, and intersects at point P.
If m∠APC = \(\frac{1}{2}\)(m\(\widehat{B D}\) – m\(\widehat{A C}\)). then point P is _________ the circle.
Answer:

Question 2.
WRITING
Explain how to find the measure of a circumscribed angle.
Answer:
A circumscribed angle is the angle made by two intersecting tangent lines to a circle. Draw lines from the circle centre to the point of tangency. The angle between the radius and tangent line is 90°. The sum of angles of a quadrilateral is 360°. Angles between radii and tangent lines is 180°. The angle at two tangent lines meet is circumscribed angle.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, line t is tangent to the circle. Find the indicated measure.

Question 3.
m\(\widehat{A B}\)

Answer:

Question 4.
m\(\widehat{D E F}\)

Answer:
m\(\widehat{D E F}\) = 2(117°) = 234°

Question 5.
m < 1

Answer:

Question 6.
m ∠ 3

Answer:
m ∠ 3 = ½ • 140 = 70°

In Exercises 7 – 14, find the value of x.

Question 7.

Answer:

Question 8.

Answer:
x° = ½ • (30 + 2x – 30)

Question 9.

Answer:

Question 10.

Answer:
34° = ½ (3x – 2 – (x + 6))
34° = ½ (3x – 2 – x – 6)
34° = ½ (2x – 8)
34° = x – 4
x° = 34 + 4
x° = 38°

Question 11.

Answer:

Question 12.

Answer:
6x – 11 = 2  • 125
6x = 250 + 11
6x = 261
x° = 43.5°

Question 13.

Answer:

Question 14.

Answer:
17x° = 75°
x° = 4.41°

ERROR ANALYSIS
In Exercises 15 and 16, describe and correct the error in finding the angle measure.

Question 15.

Answer:

Question 16.

Answer:
m∠1 = ½ (122 – 70)
= ½ (52) = 26
So, m∠1 = 26°

In Exercises 17 – 22, find the indicated angle measure. justify your answer.

Question 17.
m ∠ 1
Answer:

Question 18.
m ∠ 2
Answer:
m ∠ 2 = 60°

Explanation:
m ∠ 3 =30°, So, m ∠ 2 = 180° – (90° + 30°)
= 180° – 120° = 60°
Therefore, m ∠ 2 = 60°

Question 19.
m ∠ 3
Answer:

Question 20.
m ∠ 4
Answer:
m ∠ 4 = 90°

Question 21.
m ∠ 5
Answer:

Question 22.
m ∠ 6
Answer:
m ∠ 6 = 180° – (60° + 30° + 30°) = 180° – 120°
m ∠ 6 = 60°

Question 23.
PROBLEM SOLVING
You are flying in a hot air balloon about 1.2 miles above the ground. Find the measure of the arc that represents the part of Earth you can see. The radius of Earth is about 4000 miles.

Answer:

Question 24.
PROBLEM SOLVING
You are watching fireworks over San Diego Bay S as you sail away in a boat. The highest point the fireworks reach F is about 0.2 mile above the bay. Your eyes E are about 0.01 mile above the water. At point B you can no longer see the fireworks because of the curvature of Earth. The radius of Earth is about 4000 miles, and \(\overline{F E}\) is tangent to Earth at point T. Find m\(\widehat{s B}\). Round your answer to the nearest tenth.

Answer:

Question 25.
MATHEMATICAL CONNECTIONS
In the diagram, \(\vec{B}\)A is tangent to ⊙E. Write an algebraic expression for m\(\widehat{C D}\) in terms of x. Then find m\(\widehat{C D}\).

Answer:

Question 26.
MATHEMATICAL CONNECTIONS
The circles in the diagram are concentric. Write an algebraic expression for c in terms of a and b.

Answer:
a° = ½(c° – b°)

Question 27.
ABSTRACT REASONING
In the diagram. \(\vec{P}\)L is tangent to the circle, and \(\overline{K J}\) is a diameter. What is the range of possible angle measures of ∠LPJ? Explain your reasoning.

Answer:

Question 28.
ABSTRACT REASONING
In the diagram, \(\overline{A B}\) is an chord that is not a diameter of the circle. Line in is tangent to the circle at point A. What is the range of possible values of x? Explain your reasoning. (The diagram is not drawn to scale.)

Answer:
The possible values of x are less than 180°.

Question 29.
PROOF
In the diagram and are secant lines that intersect at point L. Prove that m∠JPN > m∠JLN.

Answer:

Question 30.
MAKING AN ARGUMENT
Your friend claims that it is possible for a circumscribed angle to have the same measure as its intercepted arc. Is your friend correct? Explain your reasoning.

Answer:
Yes, when the circumscribed angle measures 90°, the central angle measures 90°, so the intercepted arc also measures 90°.

Question 31.
REASONING
Points A and B are on a circle, and t is a tangent line containing A and another point C.
a. Draw two diagrams that illustrate this situation.
b. Write an equation for m\(\widehat{A B}\) in terms of m∠BAC for each diagram.
c. For what measure of ∠BAC can you use either equation to find m\(\widehat{A B}\)? Explain.
Answer:

Question 32.
REASONING
∆XYZ is an equilateral triangle inscribed in ⊙P. AB is tangent to ⊙P at point X, \(\overline{B C}\) is tangent to ⊙P at point Y. and \(\overline{A C}\) is tangent to ⊙P at point Z. Draw a diagram that illustrates this situation. Then classify ∆ABC by its angles and sides. Justify your answer.
Answer:

Question 33.
PROVING A THEOREM
To prove the Tangent and Intersected Chord Theorem (Theorem 10. 14), you must prove three cases.
a. The diagram shows the case where \(\overline{A B}\) contains the center of the circle. Use the Tangent Line to Circle Theorem (Theorem 10.1) to write a paragraph proof for this case.

b. Draw a diagram and write a proof for the case where the center of the circle is in the interior of ∠CAB.
c. Draw a diagram and write a proof for the case where the center of the circle is in the exterior of ∠CAB.
Answer:

Question 34.
HOW DO YOU SEE IT?
In the diagram, television cameras are Positioned at A and B to record what happens on stage. The stage is an arc of ⊙A. You would like the camera at B to have a 30° view of the stage. Should you move the camera closer or farther away? Explain your reasoning.

Answer:
25° = ½(80° – 30°) = ½(50°)
So, you should move the camera closer.

Question 35.
PROVING A THEOREM
Write a proof of the Angles Inside the Circle Theorem (Theorem 10.15).

Given Chords \(\overline{A C}\) and \(\overline{B D}\) intersect inside a circle.
Prove m ∠ 1 = \(\frac{1}{2}\)(m\(\widehat{D C}\) + m\(\widehat{A B}\))
Answer:

Question 36.
THOUGHT PROVOKING
In the figure, and are tangent to the circle. Point A is any point on the major are formed by the endpoints of the chord \(\overline{B C}\). Label all congruent angles in the figure. Justify your reasoning.

Answer:
m∠CPB = ½(CAB – CB)

Question 37.
PROVING A THEOREM
Use the diagram below to prove the Angles Outside the Circle Theorem (Theorem 10.16) for the case of a tangent and a secant. Then copy the diagrams for the other two cases on page 563 and draw appropriate auxiliary segments. Use your diagrams to prove each case.

Answer:

Question 38.
PROVING A THEOREM
Prove that the Circumscribed Angle Theorem (Theorem 10.17) follows from the Angles Outside the Circle Theorem (Theorem 10.16).
Answer:

In Exercises 39 and 40, find the indicated measure(s). justify your answer

Question 39.
Find m ∠ P when m\(\widehat{W Z Y}\) = 200°

Answer:

Question 40.
Find m\(\widehat{A B}\) and m\(\widehat{E D}\)

Answer:
m\(\widehat{E D}\) = ½ (115°) = 57.5°
∠GJA = 30°

Maintaining Mathematical Proficiency

Solve the equation.

Question 41.
x² + x = 12
Answer:

Question 42.
x² = 12x + 35

Answer:

Explanation:
x² = 12x + 35
x = \(\frac { 12 ± √(144 + 140)  }{ 2 } \)
x = \(\frac { 12 ± √284  }{ 2 } \)
x = \(\frac { 12 + √284  }{ 2 } \), \(\frac { 12 – √284  }{ 2 } \)

Question 43.
– 3 = x² + 4x
Answer:

10.6 Segment Relationships in Circles

Exploration 1

Segments Formed by Two Intersecting Chords

Work with a partner: Use dynamic geometry software.

Sample

a.
Construct two chords \(\overline{B C}\) and \(\overline{D E}\) that intersect in the interior of a circle at a point F.
Answer:

b.
Find the segment lengths BE, CF, DF, and EF and complete the table. What do you observe?

Answer:

c. Repeat parts (a) and (b) several times. Write a conjecture about your results.
REASONING ABSTRACTLY
To be proficient in math, you need to make sense of quantities and their relationships in problem situations.
Answer:

Exploration 2

Secants Intersecting Outside a Circle

Work with a partner: Use dynamic geometry software.

Sample

a. Construct two secant and that intersect at a point B outside a circle, as shown.
Answer:

b. Find the segment lengths BE, BC, BF, and BD. and complete the table. What do you observe?

Answer:

c. Repeat parts (a) and (b) several times. Write a conjecture about your results.
Answer:

Communicate Your Answer

Question 3.
What relationships exist among the segments formed by two intersecting chords or among segments of two secants that intersect outside a circle?

Question 4.
Find the segment length AF in the figure at the left.

Answer:

Explanation:
EA • AF = AD • AC
18 • AF = 9 • 8
AF = 4

Lesson 10.6 Segment Relationships in Circles

Monitoring Progress

Find the value of x.

Question 1.

Answer:
x = 8

Explanation:
4 • 6 = 3 • x
3x = 24
x = 8

Question 2.

Answer:
x = 5

Explanation:
2 • x + 1 = 4 • 3
x + 1 = 6
x = 5

Question 3.

Answer:
x = \(\frac { 54 }{ 5 } \)

Explanation:
6 • 9 = 5 • x
54 = 5x
x = \(\frac { 54 }{ 5 } \)

Question 4.

Answer:
x = \(\frac { 3 ± √37 }{ 2 } \)

Explanation:
3 • x + 2 = x + 1 • x – 1
3x + 6 = x² – 1
x² – 3x – 7 = 0
x = \(\frac { 3 ± √(9 + 28) }{ 2 } \)
x = \(\frac { 3 ± √37 }{ 2 } \)

Question 5.

Answer:
x = ±√3

Explanation:
x² = 3 • 1
x² = 3
x = ±√3

Question 6.

Answer:
x = \(\frac { 49 }{ 5 } \)

Explanation:
7² = 5 • x
49 = 5x
x = \(\frac { 49 }{ 5 } \)

Question 7.

Answer:
x = 14.4

Explanation:
12² = 10x
144 = 10x
x = 14.4

Question 8.
WHAT IF?
In Example 4, CB = 35 feet and CE = 14 feet. Find the radius of the tank.

Answer:
The radius of the tank is 36.75

Explanation:
CB² = CE ⋅ CD
35² = 14 ⋅ (2r + 14)
1225 = 28r + 196
28r = 1029
r = 36.75

Exercise 10.6 Segment Relationships in Circles

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
The part of the secant segment that is outside the circle is called a(n) ______________ .
Answer:

Question 2.
WRITING
Explain the difference between a tangent segment and a secant segment.

Answer:
A tangent segment intersects the circle at only one point. It actually doesn’t go through the circle. If a ball is rolling on a table top, then it would be the tangent. A secant segment intersects the circle in two points. It goes through the circle. In a tangent, no part is in the interior of the circle. In a secant, there is a part in the interior called a chord.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, find the value of x.

Question 3.

Answer:

Question 4.

Answer:
x = 23

Explanation:
10 • 18 = 9 • (x – 3)
20 = x – 3
x = 23

Question 5.

Answer:

Question 6.

Answer:
x = 5

Explanation:
2x • 12 = 15 • (x + 3)
24x = 15x + 45
9x = 45
x = 5

In Exercises 7 – 10, find the value of x.

Question 7.

Answer:

Question 8.

Answer:
x = \(\frac { 35 }{ 4 } \)

Explanation:
5 • 7 = 4 • x
4x = 35
x = \(\frac { 35 }{ 4 } \)

Question 9.

Answer:

Question 10.

Answer:
x = 30

Explanation:
45 • x = 50 • 27
45x = 1350
x = 30

In Exercises 11 – 14. find the value of x.

Question 11.

Answer:

Question 12.

Answer:
x = 48

Explanation:
24² = 12x
576 = 12x
x = 48

Question 13.

Answer:

Question 14.

Answer:
x = 1.5

Explanation:
3 = 2x
x = 1.5

Question 15.
ERROR ANALYSIS
Describe and correct the error in finding CD.

Answer:

Question 16.
MODELING WITH MATHEMATICS
The Cassini spacecraft is on a mission in orbit around Saturn until September 2017. Three of Saturn’s moons. Tethys. Calypso, and Teleslo. have nearly circular orbits of radius 2,95,000 kilometers. The diagram shows the positions of the moons and the spacecraft on one of Cassini’s missions. Find the distance DB from Cassini to Tethys when \(\overline{A D}\) is tangent to the circular orbit.

Answer:
BD = 579493 km

Explanation:
(203,000)² = 83000x
x = 496493
BC = 496493
BD = 496493 + 83000 = 579493

Question 17.
MODELING WITH MATHEMATICS
The circular stone mound in Ireland called Newgrange has a diameter of 250 feet. A passage 62 feet long leads toward the center of the mound. Find the perpendicular distance x from the end of the passage to either side of the mound.

Answer:

Question 18.
MODELING WITH MATHEMATICS
You are designing an animated logo for our website. Sparkles leave point C and move to the Outer circle along the segments shown so that all of the sparkles reach the outer circle at the same time. Sparkles travel from point C to point D at 2 centimeters per second. How fast should sparkles move from point C to point N? Explain.

Answer:
5.33 should sparkles move from point C to point N.

Explanation:
4 • 8 = 6 • x
x = 5.33

Question 19.
PROVING A THEOREM
Write a two-column proof of the Segments of Chords Theorem (Theorem 10.18).

Plan for Proof:
Use the diagram to draw \(\overline{A C}\) and \(\overline{D B}\). Show that ∆EAC and ∆EDB are similar. Use the fact that corresponding side lengths in similar triangles are proportional.

Answer:

Question 20.
PROVING A THEOREM
Prove the Segments of Secants Theorem (Theorem 10.19). (Hint: Draw a diagram and add auxiliary line segments to form similar triangles.)
Answer:

Question 21.
PROVING A THEOREM
Use the Tangent Line to Circle Theorem (Theorem 10. 1) to prove the Segments of Secants and Tangents Theorem (Theorem 10.20) for the special case when the secant segment Contains the center of the circle.
Answer:

Question 22.
PROVING A THEOREM
Prove the Segments of Secants and Tangents Theorem (Theorem 10.20). (Hint: Draw a diagram and add auxiliary line segments to form similar triangles.)
Answer:

Question 23.
WRITING EQUATIONS
In the diagram of the water well, AB, AD, and DE are known. Write an equation for BC using these three measurements.

Answer:

Question 24.
HOW DO YOU SEE IT?
Which two theorems would you need to use to tind PQ? Explain your reasoning.

Answer:

Question 25.
CRITICAL THINKING
In the figure, AB = 12, BC = 8, DE = 6, PD = 4, and a is a point of tangency. Find the radius of ⊙P.

Answer:

Question 26.
THOUGHT PROVOKING
Circumscribe a triangle about a circle. Then, using the points of tangency, inscribe a triangle in the circle. Must it be true that the two triangles are similar? Explain your reasoning.

Answer:

Maintaining Mathematical Proficiency

Solve the equation by completing the square.

Question 27.
x² + 4x = 45
Answer:

Question 28.
x² – 2x – 1 = 8

Answer:
x = 1 + √10, x = 1 – √10

Explanation:
x² – 2x – 1 = 8
x² – 2x – 9 = 0
x = \(\frac { 2 ± √(4 + 36) }{ 2 } \)
x = \(\frac { 2 ± √40 }{ 2 } \)
x = 1 + √10, x = 1 – √10

Question 29.
2x² + 12x + 20 = 34
Answer:

Question 30.
– 4x² + 8x + 44 = 16

Answer:
x = 1 + √8, x = 1 – √8

Explanation:
– 4x² + 8x + 44 = 16
4x² – 8x – 28 =0
x² – 2x – 7 = 0
x = \(\frac { 2 ± √(4 + 28) }{ 2 } \)
x = \(\frac { 2 ± √32 }{ 2 } \)
x = 1 ± √8

10.7 Circles in the Coordinate Plane

Exploration 1

The Equation of a Circle with Center at the Origin

Work with a partner: Use dynamic geometry software to Construct and determine the equations of circles centered at (0, 0) in the coordinate plane, as described below.

a. Complete the first two rows of the table for circles with the given radii. Complete the other rows for circles with radii of your choice.

Answer:

b. Write an equation of a circle with center (0, 0) and radius r.

Answer:
x²  + y²  = r²

Explanation:
(x – 0)² + (y – 0)²  = r²
x²  + y²  = r²

Exploration 2

The Equation of a Circle with Center (h, k)

Work with a partner: Use dynamic geometry software to construct and determine the equations of circles of radius 2 in the coordinate plane, as described below.

a. Complete the first two rows of the table for circles with the given centers. Complete the other rows for circles with centers of your choice.

Answer:

b. Write an equation of a circle with center (h, k) and radius 2.

Answer:
(x – h)² + (y – k)² = 4

c. Write an equation of a circle with center (h, k) and radius r.

Answer:
(x – h)² + (y – k)² = r²

Exploration 3

Deriving the Standard Equation of a Circle

Work with a partner. Consider a circle with radius r and center (h, k).

Write the Distance Formula to represent the distance d between a point (x, y) on the circle and the center (h, k) of the circle. Then square each side of the Distance Formula equation.

How does your result compare with the equation you wrote in part (c) of Exploration 2?

MAKING SENSE OF PROBLEMS
To be proficient in math, you need to explain correspondences between equations and graphs.

Answer:
(x – h)² + (y – k)² = r²

Communicate Your Answer

Question 4.
What is the equation of a circle with center (h, k) and radius r in the coordinate plane?

Answer:
(x – h)² + (y – k)² = r²

Question 5.
Write an equation of the circle with center (4, – 1) and radius 3.

Answer:
x² + y² – 8x + 2y + 8 = 0

Explanation:
(x – 4)² + (y + 1)² = 9
x² – 8x + 16 + y² + 2y + 1 = 9
x² + y² – 8x + 2y = 9 – 17
x² + y² – 8x + 2y + 8 = 0

Lesson 10.7 Circles in the Coordinate Plane

Monitoring Progress

Write the standard equation of the circle with the given center and radius.

Question 1.
center: (0, 0), radius: 2.5

Answer:
x² + y² = 6.25

Explanation:
(x – 0)² + (y – 0)² = 2.5²
x² + y² = 6.25

Question 2.
center: (- 2, 5), radius: 7

Answer:
(x + 2)² + (y – 5)² = 49

Explanation:
(x + 2)² + (y – 5)² = 7²
(x + 2)² + (y – 5)² = 49

Question 3.
The point (3, 4) is on a circle with center (1, 4). Write the standard equation of the circle.

Answer:
(x – 1)² + (y – 4)² = 4

Explanation:
r = √(3 – 1)² + (4 – 4)²
= √(2)²
r = 2
(x – 1)² + (y – 4)² = 2²
(x – 1)² + (y – 4)² = 4

Question 4.
The equation of a circle is x² + y² – 8x + 6y + 9 = 0. Find the center and the radius of the circle. Then graph the circle.

Answer:
The center of the circle (4, -3) and radius is 4.

Explanation:
x² + y² – 8x + 6y + 9 = 0
x² – 8x + 16 + y² + 6y + 9 = 16
(x – 4)² + (y + 3)² = 4²
The center of the circle (4, -3) and radius is 4.

Question 5.
Prove or disprove that the point (1, √5 ) lies on the circle centered at the origin and containing the point (0, 1).

Answer:
Disproved.

Explanation:
We consider the circle centred at the origin and containing the point (0, 1).
Therefore, we can conclude that the radius of the circle r = 1, let O (0, 0) and B (1, √5). So the distance between two points is
OB = √(1 – 0) + (√5 – 0)² = √(1 + 5) = √6
As the radius of the given circle is 1 and distance of the point B from its centre is √6. So we can conclude that point does lie on the given circle.

Question 6.
why are three seismographs needed to locate an earthquake’s epicentre?

Answer:

Exercise 10.7 Circles in the Coordinate Plane

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
What is the standard equation of a circle?
Answer:

Question 2.
WRITING
Explain why knowing the location of the center and one point on a circle is enough to graph the circle.

Answer:
If we know the location of the center and one point on the circle, we can graph a circle because the distance from the center to the point is called the radius.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 8, write the standard equation of the circle.

Question 3.

Answer:

Question 4.

Answer:
x² + y² = 36

Explanation:
The center is (0, 0) and the radius is 6
(x – h)² + (y – k)² = r²
(x – 0)² + (y – 0)² = 6²
x² + y² = 36

Question 5.
a circle with center (0, 0) and radius 7
Answer:

Question 6.
a circle with center (4, 1) and radius 5

Answer:
(x – 4)² + (y – 1)² = 25

Explanation:
(x – h)² + (y – k)² = r²
(x – 4)² + (y – 1)² = 5²
(x – 4)² + (y – 1)² = 25

Question 7.
a circle with center (- 3, 4) and radius 1
Answer:

Question 8.
a circle with center (3, – 5) and radius 7

Answer:
(x – 3)² + (y + 5)² = 49

Explanation:
(x – h)² + (y – k)² = r²
(x – 3)² + (y + 5)² = 7²
(x – 3)² + (y + 5)² = 49

In Exercises 9 – 11, use the given information to write the standard equation of the circle.

Question 9.
The center is (0, 0), and a point on the circle is (0, 6).
Answer:

Question 10.
The center is (1, 2), and a point on the circle is (4, 2).

Answer:
x² + y² = 9

Explanation:
r = √(x – h)² + (y – k)²
= √(4 – 1)² + (2 – 2)²
= √3²
r = 3
(x – h)² + (y – k)² = r²
(x – 0)² + (y – 0)² = 3²
x² + y² = 9

Question 11.
The center is (0, 0). and a point on the circle is (3, – 7).
Answer:

Question 12.
ERROR ANALYSIS
Describe and correct the error in writing the standard equation of a circle.

Answer:
(x – h)² + (y – k)² = r²
(x + 3)² + (y + 5)² = 3²
(x + 3)² + (y + 5)² = 9

In Exercises 13 – 18, find the center and radius of the circle. Then graph the circle.

Question 13.
x² + y² = 49
Answer:

Question 14.
(x + 5)² + (y – 3)² = 9

Answer:
Center is (-5, 3) and rdaius is 3.

Explanation:
For the equation (x + 5)² + (y – 3)² = 9, center is (-5, 3) and rdaius is 3.

Question 15.
x² + y² – 6x = 7
Answer:

Question 16.
x² + y² + 4y = 32

Answer:
The center is (0, -2) and radius is 6

Explanation:
x² + y² + 4y = 32
x² + y² + 4y + 4 = 32 + 4
x² + (y + 2)² = 36
(x – 0)² + (y – (-2))² = 6²
The center is (0, -2) and radius is 6

Question 17.
x² + y² – 8x – 2y = – 16
Answer:

Question 18.
x² + y² + 4x + 12y = – 15

Answer:
The center is (-2, -6) and radius is 5

Explanation:
x² + y² + 4x + 12y = – 15
x² + 4x + 4 + y² + 12y + 36 = -15 + 36 + 4
(x + 2)² + (y + 6)² = 5²
The center is (-2, -6) and radius is 5

In Exercises 19 – 22, prove or disprove the statement.

Question 19.
The point (2, 3) lies on the circle centered at the origin with radius 8.
Answer:

Question 20.
The point (4, √5) lies on the circle centered at the origin with radius 3.

Answer:
The point (4, √5) does not lie on the circle.

Explanation:
r² = (x – h)² + (y – k)²
3² ______________ (4 – 0)² + (√5 – 0)²
9 ______________ 16 + 5
9 ≠ 21
Because the radius is 3 and the distance between center and the point is more than the radius. So the point does not lie on the circle.

Question 21.
The point (√6, 2) lies on the circle centered at the origin and containing the point (3, – 1).
Answer:

Question 22.
The point (√7, 5) lies on the circle centered at the origin and containing the point (5, 2).

Answer:
The point (√7, 5) does not lie on the circle.

Explanation:
r² = (x – h)² + (y – k)²
= (√7 – 0)² + (5 – 0)² = 7 + 25 = 32
r = 5.65
(5.65)² ______________ (5 – 0)² + (2 – 0)²
32 ______________ 25 + 4
32 ≠ 29
Because the radius is 3 and the distance between center and the point is more than the radius. So the point does not lie on the circle.

Question 23.
MODELING WITH MATHEMATICS
A City’s commuter system has three zones. Zone I serves people living within 3 miles of the city’s center. Zone 2 serves those between 3 and 7 miles from the center. Zone 3 serves those over 7 miles from the center.

a. Graph this Situation on a coordinate plane where each unit corresponds to 1 mile. Locate the city’s center at the origin.
b. Determine which zone serves people whose homes are represented by the points (3, 4), (6, 5), (1, 2), (0.3). and (1, 6).
Answer:

Question 24.
MODELING WITH MATHEMATICS
Telecommunication towers can be used to transmit cellular phone calls. A graph with units measured in kilometers shows towers at points (0, 0), (0, 5), and (6, 3). These towers have a range of about 3 kilometers.
a. Sketch a graph and locate the towers. Are there any locations that may receive calls from more than one tower? Explain your reasoning.
Answer:

b. The center of City A is located at (- 2, 2.5), and the center of City B is located at (5, 4). Each city has a radius of 1.5 kilometers. Which city seems to have better cell phone coverage? Explain your reasoning.

Answer:
There are three towers at points (0, 0), (0, 5) and (6, 3) with range of about 3 km
a. Let’s sketch the graph to locate towers. Draw the points (0, 0), (0, 5) and (6, 3). Then draw three circles with centers at these points and radii 3.
There are locations that can receive calls from more that one tower because circles with centers (0, 0) and (0, 5) overlap. Locations in their intersection can receive calls from two towers.
The city A has a center located at (-2, 2.5) and city B has a center located at (5, 4). Both cities have radius 1.5 km
Let’s draw the city A as a circle with center (-2, 2.5) and radius 1.5 and city B with center (5, 4) and radius 1.5.
From the graph we can conclude that the city B has better cell phone coverage because parts of city A do not have coverage.

Question 25.
REASONING
Sketch the graph of the circle whose equation is x² + y² = 16. Then sketch the graph of the circle after the translation (x, y) → (x – 2, y – 4). What is the equation of the image? Make a conjecture about the equation of the image of a circle centered at the origin after a translation m units to the left and n units down.
Answer:

Question 26.
HOW DO YOU SEE IT?
Match each graph with its equation.

a.	A. x2 + (y + 3) 2 = 4
b.	B. (x – 3) 2 + y2 = 4
c.	C. (x + 3) 2 + y2 = 4
d.	D. x2 + (y – 3) 2 = 4

Answer:
a ➝ C, b ➝ A, c ➝ D, d ➝ B

Question 27.
USING STRUCTURE
The vertices of ∆XYZ are X(4, 5), Y(4, 13), and Z(8, 9). Find the equation of the circle circumscribed about ∆XYZ. Justify your answer.
Answer:

Question 28.
THOUGHT PROVOKING
A circle has center (h, k) and contains point (a, b). Write the equation of the line tangent to the circle at point (a, b).

Answer:
y – b = \(\frac { h – a }{ b – k } \)(x – a)

Explanation:
It is given a circle with center C(h, k. A circle with point A(a, b). We have to write the equation of a tangent that intersects the circle at point A
By the tangent line to circle theorem, a tangent is perpendicular to the radius. Two lines are perpendicular if and only if their slopes are negative reciprocals. So, find the equation of the line AC to know its slope.
The equation of the line which has two points (a, b), (c, d) is y – b = \(\frac { b – c }{ a – d } \)(x – a)
The equation of the line which has two points A(a, b) and C(h, k) is
y – b = \(\frac { b – k }{ a – h } \)(x – a)
Therefore, the slope of the line through A nad C is \(\frac { b – k }{ a – h } \)
Hence the slope of the tangent is –\(\frac { a – h }{ b – k } \) = \(\frac { h – a }{ b – k } \)
Use the equation of the line y = kx + n through the point (a, b)
y – b = k(x – a) to find the equation of the tangent
The equation of the tangent with slope \(\frac { h – a }{ b – k } \) and through the point A(a, b) is y – b = \(\frac { h – a }{ b – k } \)(x – a)

MATHEMATICAL CONNECTIONS
In Exercises 29 – 32, use the equations to determine whether the line is a tangent, a secant a secant that contains the diameter, or name of these. Explain your reasoning.

Question 29.
Circle: (x – 4)² + (y – 3)² = 9
Line: y = 6
Answer:

Question 30.
Circle: (x + 2)² + (y – 2)² = 16
Line: y = 2x – 4

Answer:
The line is a secant line.

Explanation:
(x + 2)² + (y – 2)² = 16, y = 2x – 4
(x + 2)² + (2x – 4 – 2)² = 16
x² + 4x + 4 + (2x – 6)² = 16
x² + 4x + 4 + 4x² – 24x + 36 = 16
5x² – 20x + 40 – 16 = 0
5x² – 20x + 24 = 0
x = \(\frac { 20 ±√-80 }{ 10 } \)
x = 2, y = 2 • 2 – 4 = 0, (2, 0)
The system has two solutions and point does not lie on the line. So the line is a secant line.

Question 31.
Circle: (x – 5)² + (y + 1)² = 4
Line: y = \(\frac{1}{5}\)x – 3\
Answer:

Question 32.
Circle: (x + 3)² + (y – 6)² = 25
Line: y = –\(\frac{4}{3}\)x + 2

Answer:
The line is a secant line.

Explanation:
(x + 3)² + (y – 6)² = 25, y = –\(\frac{4}{3}\)x + 2
(x + 3)² + (-\(\frac{4}{3}\)x + 2 – 6)² = 25
x² + 6x + 9 + \(\frac { 16x² }{ 9 } \) + \(\frac { 32x }{ 3 } \) + 16 = 25
\(\frac { 25x² }{ 9 } \) + \(\frac { 50x }{ 3 } \) = 0
x(25x + 150) = 0
x = 0 or x = -6
y = 2, y = 10
(0, 2) and (-6, 10)
d = √(0 + 6)² + (2 – 10)²
= √(36 + 64)
= 10 ≠ 5
The system has two solutions and point does not lie on the line. So the line is a secant line.

MAKING AN ARGUMENT
Question 33.
Your friend claims that the equation of a circle passing through the points (- 1, 0) and (1, 0) is x² – 2yk + y² = 1 with center (0, k). Is your friend correct? Explain your reasoning.
Answer:

Question 34.
REASONING
Four tangent circles are centered on the x-axis. The radius of ⊙A is twice the radius of ⊙O, The radius of ⊙B is three times the radius of ⊙O, The radius of ⊙C is four times the radius of ⊙O, All circles have integer radii, and the point (63, 16) is On ⊙C. What is the equation of ⊙A? Explain your reasoning.

Answer:

Maintaining Mathematical Proficiency

Identify the arc as a major arc, minor arc, or semicircle. Then find the measure of the arc.

Question 35.
\(\widehat{R S}\)
Answer:

Question 36.
\(\widehat{P R}\)

Answer:
\(\widehat{P R}\) is a right angle
\(\widehat{P R}\) = 25 + 65 = 90°

Question 37.
\(\widehat{P R T}\)
Answer:

Question 38.
\(\widehat{S T}\)

Answer:
\(\widehat{S T}\) is a major arc
\(\widehat{S T}\) = 360 – (90 + 65 +25 + 53) = 127°

Question 39.
\(\widehat{R S T}\)
Answer:

Question 40.
\(\widehat{Q S}\)

Answer:
\(\widehat{Q S}\) is a minor arc
\(\widehat{Q S}\) = 25 + 53 = 78°

Circles Review

10.1 Lines and Segments That Intersect Circles

Tell whether the line, ray, or segment is best described as a radius, chord, diameter, secant, or tangent of ⊙P.

Question 1.
\(\overline{P K}\)

Answer:
\(\overline{P K}\) is radius

Question 2.
\(\overline{N M}\)

Answer:
\(\overline{N M}\) is chord

Question 3.
\(\vec{J}\)L

Answer:
\(\vec{J}\)L is tangent

Question 4.
\(\overline{K N}\)

Answer:
\(\overline{K N}\) is diameter

Question 5.

Answer:
NL is secant

Question 6.
\(\overline{P N}\)

Answer:
\(\overline{P N}\) is radius

Tell whether the common tangent is internal or external.

Question 7.

Answer:
Internal common tangent

Question 8.

Answer:
External common tangent

Points Y and Z are points of tangency. Find the value of the variable.

Question 9.

Answer:
a = \(\frac { 3 ± 33 }{ 18 } \)

Explanation:
3a = 9a² – 30
9a² – 3a – 30 = 0
a = \(\frac { 3 ± 33 }{ 18 } \)

Question 10.

Answer:
c = 2

Explanation:
2c² + 9c + 6 = 9c + 14
2c² – 8 = 0
c² – 4 = 0
c = 2

Question 11.

Answer:
r = 12

Explanation:
(3 + r)² = r² + 9²
9 + 6r + r² = r² + 81
6r = 72
r = 12

Question 12.
Tell whether \(\overline{B D}\) is tangent to ⊙C. Explain.

Answer:
\(\overline{B D}\) is not tangent to ⊙C

Explanation:
52² = 10² + 48²
2704 = 100 + 2304
So, \(\overline{B D}\) is not tangent to ⊙C

10.2 Finding Arc Measures

Use the diagram above to find the measure of the indicated arc.

Question 13.
\(\widehat{K L}\)

Answer:
\(\widehat{K L}\) = 100°

Explanation:
\(\widehat{K L}\) = ∠KPL = 100°

Question 14.
\(\widehat{L M}\)

Answer:
\(\widehat{L M}\) = 60°

Explanation:
\(\widehat{L M}\) = 180° – 120°
= 60°

Question 15.
\(\widehat{K M}\)

Answer:
\(\widehat{K M}\) = 160°

Explanation:
\(\widehat{K M}\) = 100° + 60°
= 160°

Question 16.
\(\widehat{K N}\)

Answer:
\(\widehat{K N}\) = 80°

Explanation:
\(\widehat{K N}\) = 360 – (120 + 100 + 60)
= 360 – 280 = 80°

Tell whether the red arcs are congruent. Explain why or why not.

Question 17.

Answer:
\(\widehat{S T}\), \(\widehat{Y Z}\) are not congruent.

Explanation:
\(\widehat{S T}\), \(\widehat{Y Z}\) are not congruent. Because both arcs are from different circles and having different radii.

Question 18.

Answer:
\(\widehat{A B}\), \(\widehat{E F}\) are congruent.

Explanation:
\(\widehat{A B}\), \(\widehat{E F}\) are congruent. Because those circles have same radii.

10.3 Using Chords

Find the measure of \(\widehat{A B}\).

Question 19.

Answer:
\(\widehat{A B}\) = 61°

Explanation:
\(\widehat{A B}\) = 61°
If ED = AB, then \(\widehat{A B}\) = \(\widehat{E D}\)

Question 20.

Answer:
\(\widehat{A B}\) = 65°

Explanation:
\(\widehat{A B}\) = \(\widehat{A D}\)
So, \(\widehat{A B}\) = 65°

Question 21.

Answer:
\(\widehat{A B}\) = 91°

Explanation:
\(\widehat{A B}\) = \(\widehat{E D}\)
So, \(\widehat{A B}\) = 91°

Question 22.
In the diagram. QN = QP = 10, JK = 4x, and LM = 6x – 24. Find the radius of ⊙Q.

Answer:
The radius of ⊙Q is 26.

Explanation:
6x – 24 = 4x
6x – 4x = 24
2x = 24
x = 12
ML = 6(12) – 24 = 48
JN = \(\frac { 48 }{ 2 } \) = 24
JQ² = JN² + NQ²
= 24² + 10² = 576 + 100
JQ = 26
Therefore, the radius of ⊙Q is 26

10.4 Inscribed Angles and Polygons

Find the value(s) of the variable(s).

Question 23.

Answer:
x° = 80°

Explanation:
x° = 2 • 40° = 80°

Question 24.

Answer:
q° = 100°, r° = 20°

Explanation:
q° + 80° = 180°
q° = 100°
4r° + 100 = 180°
4r° = 80°
r° = 20°

Question 25.

Answer:
d° = 5°

Explanation:
14d° = 70°
d° = 5°

Question 26.

Answer:
y° = 30°, z° = 10°

Explanation:
3y° = 90°
y° = 30°
50° + 90° + 4z° = 180°
4z° = 40°
z° = 10°

Question 27.

Answer:
m° = 44°
n° = 39°

Explanation:
m° = 44°
n° = 39°

Question 28.

Answer:
c° = 28°

Explanation:
c° = ½ • 56 = 28

10.5 Angle Relationships in Circles

Find the value of x.

Question 29.

Answer:
x° = 250°

Explanation:
x° = 250°

Question 30.

Answer:
x° = 106°

Explanation:
x° = ½(152 + 60)
= ½(212) = 106°

Question 31.

Answer:
x° = 28°

Explanation:
x° = ½(96 – 40)
= ½(56) = 28°

Question 32.
Line l is tangent to the circle. Find m\(\widehat{X Y Z}\).

Answer:
m\(\widehat{X Y Z}\) = 240°

Explanation:
m\(\widehat{X Y Z}\) = 2(120)
= 240°

10.6 Segment Relationships in Circles

Find the value of x.

Question 33.

Answer:
x = 8

Explanation:
3 • x = 4 • 6
x = 8

Question 34.

Answer:
x = 3

Explanation:
(x + 3) • x = (6 – x) • 2x
x + 3 = 12 – 2x
3x = 9
x = 3

Question 35.

Answer:
x = 18

Explanation:
12² = 8 • x
144 = 8x
x = 18

Question 36.
A local park has a circular ice skating rink. You are standing at point A, about 12 feet from the edge of the rink. The distance from you to a point of tangency on the rink is about 20 feet. Estimate the radius of the rink.

Answer:
Estimated radius of the rink is 10 ft

Explanation:
20² = 12 • (2r + 12)
400 = 24r + 144
256 = 24r
r = 10.66
Therefore, estimated radius of the rink is 10 ft

10.7 Circles in the Coordinate Plane

Write the standard equation of the circle shown.

Question 37.

Answer:
(x – 4)² + (y + 1)² = 12.25

Explanation:
(x – 4)² + (y + 1)² = 3.5²
(x – 4)² + (y + 1)² = 12.25

Question 38.

Answer:
(x – 8)² + (y – 5)² = 36

Explanation:
(x – 8)² + (y – 5)² = 6²
(x – 8)² + (y – 5)² = 36

Question 39.

Answer:
x² + y² = 4

Explanation:
(x – 0)² + (y – 0)² = 2²
x² + y² = 4

Write the standard equation of the circle with the given center and radius.

Question 40.
center: (0,0), radius: 9

Answer:
x² + y² = 81

Explanation:
(x – 0)² + (y – 0)² = 9²
x² + y² = 81

Question 41.
center: (- 5, 2), radius: 1.3

Answer:
(x + 5)² + (y – 2)² = 1.69

Explanation:
(x + 5)² + (y – 2)² = 1.3²
(x + 5)² + (y – 2)² = 1.69

Question 42.
center: (6, 21), radius: 4

Answer:
(x – 6)² + (y – 21)² = 16

Explanation:
(x – 6)² + (y – 21)² = 4²
(x – 6)² + (y – 21)² = 16

Question 43.
center: (- 3, 2), radius: 16

Answer:
(x + 3)² + (y – 2)² = 256

Explanation:
(x + 3)² + (y – 2)² = 16²
(x + 3)² + (y – 2)² = 256

Question 44.
center: (10, 7), radius: 3.5

Answer:
(x – 10)² + (y – 7)² = 12.25

Explanation:
(x – 10)² + (y – 7)² = 3.5²
(x – 10)² + (y – 7)² = 12.25

Question 45.
center: (0, 0), radius: 5.2

Answer:
x² + y² = 27.04

Explanation:
(x – 0)² + (y – 0)² = 5.2²
x² + y² = 27.04

Question 46.
The point (- 7, 1) is on a circle with center (- 7, 6). Write the standard equation of the circle.

Answer:
(x + 7)² + (y – 6)² = 25

Explanation:
r² = (-7 + 7)² + (6 – 1)²
= 5²
r = 5
And, centre is (-7, 6)
The standard equation of a circle is (x – (-7))² + (y – 6)² = 5²
(x + 7)² + (y – 6)² = 25

Question 47.
The equation of a circle is x² + y² – 12x + 8y + 48 = 0. Find the center and the radius of the circle. Then graph the circle.

Answer:
The radius of the circle is 2, the centre is (6, -4)

Explanation:
x² + y² – 12x + 8y + 48 = 0
x² – 12x + 36 + y² + 8y + 16 = 4
(x – 6)² + (y + 4)² = 2²
So, the radius of the circle is 2, the centre is (6, -4)

Question 48.
Prove or disprove that the point (4, – 3) lies on the circle centred at the origin and containing
the point (- 5, 0).

Answer:
The point (4, – 3) lies on the circle.

Explanation:
Use the distance formula to find the radius of the circle with cente (0, 0) and a point (-5, 0)
r = √(-5 – 0)² + (0 – 0)² = 5
The distance from the point (4, -3) to the center (0, 0)
d = √(4 – 0)² + (-3 – 0)² = √(16 +9) = 5
Since the radius of the circle is 5, the point lies on the circle.

Circles Chapter Test

Find the measure of each numbered angle in ⊙P. Justify your answer.

Question 1.

Answer:
m∠1 = 72.5°
m∠2 = 145°

Explanation:
m∠1 = \(\frac { 145 }{ 2 } \)
= 72.5°
m∠2 = 145°

Question 2.

Answer:
m∠1 = 60°, m∠2 = 90°

Explanation:
A tangent is perpendicualr to diameter. So, m∠2 = 90°
m∠1 = 60°

Question 3.

Answer:
m∠1 = 48°

Explanation:
m∠1 = \(\frac { 96 }{ 2 } \) = 48°

Question 4.

Answer:

Use the diagram.

Question 5.
AG = 2, GD = 9, and BG = 3. Find GF.

Answer:

Question 6.
CF = 12, CB = 3, and CD = 9. Find CE.

Answer:

Question 7.
BF = 9 and CB = 3. Find CA

Answer:

Question 8.
Sketch a pentagon inscribed in a circle. Label the pentagon ABCDE. Describe the relationship between each pair of angles. Explain your reasoning.

a. ∠CDE and ∠CAE

Answer:

b. ∠CBE and ∠CAE

Answer:

Find the value of the variable. Justify your answer.

Question 9.

Answer:
x = 5

Explanation:
5x – 4 = 3x + 6
5x – 3x = 6 + 4
2x = 10
x = 5

Question 10.

Answer:
r = 9

Explanation:
(6 + r)² = 12² + r²
36 + 12r + r² = 144 + r²
12r = 108
r = 9

Question 11.
Prove or disprove that the point (2√2, – 1) lies on the circle centered at (0, 2) and containing the point (- 1, 4).

Answer:
Disproved

Explanation:
We consider the circle centred at the A(0, 2) and containing the point B(-1, 4).
AB = √(-1 – 0)² + (4 – 2)² = √1 + 4 = √5
The distance between centre A(0, 2) and P(2√2, – 1) is
AP = √(2√2 – 0)² + (-1 – 2)² = √8 + 9 = √17
AB ≠ AP
So, the point (2√2, – 1) dies not lie on the circle.

Prove the given statement.

Question 12.
\(\widehat{S T} \cong \widehat{R Q}\)

Answer:

Question 13.
\(\widehat{J M} \cong \widehat{L M}\)

Answer:

Question 14.
\(\widehat{D G} \cong \widehat{F G}\)

Answer:

Question 15.
A bank of lighting hangs over a stage. Each light illuminates a circular region on the stage. A coordinate plane is used to arrange the lights, using a corner of the stage as the origin. The equation (x – 13)² + (y – 4)² = 16 represents the boundary of the region illuminated by one of the lights. Three actors stand at the points A(11, 4), B(8, 5), and C(15, 5). Graph the given equation. Then determine which actors are illuminated by the light.

Answer:
The equation (x – 13)² + (y – 4)²= 16 represents the standard equation of the circle with center (13, 4) and radius 4
Graph the circle with center S(13, 4), radius 4. Then graph the points A(11,4), B (8, 5), C(15,5) which represents the places where the actors stand.

From the graph, we can see that points A and C inside the circle and point B is outside the circle. Therefore, actors who stand at points A and C are illuminated by the light

Question 16.
If a car goes around a turn too quickly, it can leave tracks that form an arc of a circle. By finding the radius of the circle, accident investigators can estimate the speed of the car.
Answer:

a. To find the radius, accident investigators choose points A and B on the tire marks. Then the investigators find the midpoint C of \(\overline{A B}\). Use the diagram to find the radius r of the circle. Explain why this method works.

Answer:
The radius r of the circle = 155.71 ft

Explanation:
Given that,
AC = 130 ft, CD = 70 ft
CE = (r – 70) ft
r² = a² + b²
r²= 130²+ (r – 70)²
r² = 16900 + r² – 140r + 4900
140r = 21,800
r = 155.71 ft

b. The formula S = 3.87√fr can be used to estimate a car’s speed in miles per hour, where f is the coefficient of friction and r is the radius of the circle in feet. If f = 0.7, estimate the car’s speed in part (a).

Answer:
The estimated car’s speed is 39.67 miles per hour

Explanation:
S = 3.87√fr
S = 3.8 √(0.7 x 155.71)
S = 3.8 √108.997
S = 3.8 x 10.44
S = 39.67

Circles Cumulative Assessment

Question 1.
Classify each segment as specifically as possible.

a. \(\overline{B G}\)

Answer:
\(\overline{B G}\) is a chord

b. \(\overline{C D}\)

Answer:
\(\overline{C D}\) is radius.

c. \(\overline{A D}\)

Answer:
\(\overline{A D}\) is diameter.

d. \(\overline{F E}\)

Answer:
\(\overline{F E}\) is a chord

Question 2.
Copy and complete the paragraph proof.

Given Circle C with center (2, 1) and radius 1,
Circle D with center (0, 3) and radius 4
Prove Circle C is similar to Circle D.

Map Circle C to Circle C’ by using the _________ (x, y) → _________ so that Circle C’ and Circle D have the same center at (____, _____). Dilate Circle C’ using a cellIer of dilation (_____, _____) and a scale factor of _____ . Because there is a _________ transformation that maps Circle C to Circle D, Circle C is __________ Circle D.

Answer:
Map Circle C to Circle C’ by using the scale factor (x, y) → (0, 3) so that Circle C’ and Circle D have the same center at (2, 1). Dilate Circle C’ using a cellIer of dilation (2, 1) and a scale factor of circles. Because there is a transformation that maps Circle C to Circle D, Circle C is similar to Circle D.

Question 3.
Use the diagram to write a proof.

Given ∆JPL ≅ ∆NPL
\(\overline{P K}\) is an altitude of ∆JPL
\(\overline{P M}\) is an altitude ∆NPL
Prove ∆PKL ~ ∆NMP

Answer:
∆JPL is similar to ∆NPL and PK is the altitude of ∆JPL and PM is an altitude of ∆NPL
Altitude is a line drawn from one vertex to the opposite site. It is perpendicular to the side.
So, ∆PKL is similar to ∆NMP

Question 4.
The equation of a circle is x² + y² + 14x – 16y + 77 = 0. What are the center and radius of the circle?
(A) center: (14, – 16). radius: 8.8
(B) center: (- 7, 8), radius: 6
(C) center (- 14, 16), radius: 8.8
(D) center: (7, – 8), radius: 5.2

Answer:
(B) center: (- 7, 8), radius: 6

Explanation:
x² + y² + 14x – 16y + 77 = 0
x² + 14x + 49 + y² – 16y + 64 = 36
(x + 7)² + (y – 8)² = 6²
So, the centre is (-7, 8) and radius is 6.

Question 5.
The coordinates of the vertices of a quadrilateral are W(- 7, – 6), X(1, – 2), Y(3, – 6) and Z(- 5, – 10). Prove that quadrilateral WXYZ is a rectangle.

Answer:
Proved

Explanation:
Find the distance of WY and ZX
WY = √(-7 – 3)² + (-6 + 6)² = √(-10)² = 10
ZX = √(1 + 5)² + (-2 + 10)² = √6² + 8² = 10
WY = ZX, the diagonals are congruent
Use the slope formula to find the slopes of diagonals
Slope of WY = \(\frac { -6 + 6 }{ -7 – 3 } \) = 0
Slope of ZX = \(\frac { -2 + 10 }{ 1 + 5 } \) = \(\frac { 4 }{ 3 } \)
Because the product of slopes of diagonals is 0, the diagonals are not perpendicular
Therefore, the quadrilateral WXYZ is a rectangle.

Question 6.
Which angles have the same measure as ∠ACB? Select all that apply.
im – 295
∠DEF ∠JGK ∠KGL ∠LGM ∠MGJ
∠QNR ∠STV ∠SWV ∠VWU ∠XYZ
Answer:
∠VWU

Question 7.
Classify each related conditional statement based on the conditional statement
“If you are a soccer player. then you are an athlete.”
a. If you are not a soccer player, then you are not an athlete.

Answer:
False

b. If you are an athlete, then you are a soccer player.

Answer:
False

c. You are a soccer player if and only if you are an athlete.

Answer:
True

d. If you are not an athlete, then you are not a soccer player.

Answer:
False

Question 8.
Your friend claims that the quadrilateral shown can be inscribed in a circle. Is your friend correct? Explain our reasoning.

Answer:
If the sum of any two angles is 180°, then the quadrilateral is inscribed in a circle.
So, 70° + 110° = 180°, 110° + 70°= 180°
So, my friend is correct.