Tag: Eureka Math Geometry

Engage NY Eureka Math Geometry Module 2 Lesson 25 Answer Key

Eureka Math Geometry Module Lesson 25 Exercise Answer Key

Use the right triangle ∆ ABC to answer Exercises 1 – 3.

Exercise 1.
Name the side of the triangle opposite ∠A.
Answer:
\(\overline{B C}\)

Exercise 2.
Name the side of the triangle opposite ∠B.
Answer:
\(\overline{A C}\)

Exercise 3.
Name the side of the triangle opposite ∠C.
Answer:
\(\overline{A B}\)

For each exercise, label the appropriate sides as adjacent, opposite, and hypotenuse, with respect to the marked acute angle.

Exercise 4.

Answer:

Exercise 5.

Answer:

Exercise 6.

Answer:

For each question, round the unknown lengths appropriately. Refer back to your completed chart from the ExploratoryChallenge; each indicated acute angle is the same approximated acute angle measure as in the chart. Set up and label the appropriate length ratios, using the terms app, adj, and hyp in the setup of each ratio.

Exercise 7.

Answer:

Exercise 8.

Answer:

Exercise 9.

Answer:

Exercise 10.
From a point 120 m away from a building, Serena measures the angle between the ground and the top of a building and finds it measures 41°.
What is the height of the building? Round to the nearest meter.

Answer:

The height of the building is approximately 106 meters.

Eureka Math Geometry Module Lesson 25 Problem Set Answer Key

The table below contains the values of the ratios \(\frac{o p p}{h y p}\) and \(\frac{\text { adj }}{\text { hyp }}\) for a variety of right triangles based on a given acute angle, θ, from each triangle. Use the table and the diagram of the right triangle below to complete each problem.

Answer:

For each problem, approximate the unknown lengths to one decimal place. Write the appropriate length ratios using the terms opp, adj, and hyp in the setup of each ratio.

Question 1.
Find the approximate length of the leg opposite the 80° angle.

Answer:
Using the value of the ratio for an \(\frac{\text { opp }}{\text { hyp }}\) 80° angle:
\(\frac{\text { opp }}{24}\) = 0.9848
opp = 23. 6352
The length of the leg opposite the 80° angle is approximately 23.6.

Question 2.
Find the approximate length of the hypotenuse.

Answer:
Using the value of the ratio \(\frac{\text { opp }}{\text { hyp }}\) for a 45° angle:
\(\frac{7.1}{\text { hyp }}\) = 0.7071
\(\frac{7.1}{0.7071}\) = hyp
10.0410 ≈ hyp
The length of the hypotenuse is approximately 10.0.

Question 3.
Find the approximate length of the hypotenuse.

Answer:
adj
Using the value of the ratio \(\frac{\text { adj }}{\text { hyp }}\) for a 60° angle:
\(\frac{0.7}{\text { hyp }}\) = \(\frac{1}{2}\)
\(\frac{0.7}{\frac{1}{2}}\) = hyp
1.4 = hyp
The length of the hypotenuse is 1.4.

Question 4.
Find the approximate length of the leg adjacent to the 40° angle.

Answer:
Using the value of the ratio \(\frac{\text { adj }}{\text { hyp }}\) for a 40° angle:
\(\frac{\text { adj }}{18}\) = 0.7660
adj = 13. 788
The length of the leg adjacent to the given 40° angle is approximately
13.8.

Question 5.
Find the length of both legs of the right triangle below. Indicate which leg is adjacent and which is opposite the given angle of 30°.

Answer:
Using the value of the ratio \(\frac{\text { opp }}{\text { hyp }}\) for a 30° angle:
\(\frac{\text { opp }}{12}=\frac{1}{2}\)
opp = 6
The length of the leg that is opposite the given 30° angle is 6.

Using the value of the ratio \(\frac{\text { adj }}{\text { hyp }}\) for a 30° angle:
\(\frac{\text { adj }}{12}\) = 0.8660
adj = 10. 392
The length of the leg that is adjacent to the given 30° angle is approximately 10.4.

Question 6.
Three city streets form a right triangle. Main Street and State Street are perpendicular. Laura Street and State Street intersect at a 50° angle. The distance along Laura Street to Main Street is 0.8 mile. If Laura Street is closed between Main Street and State Street for a festival, approximately how far (to the nearest tenth) will someone have to travel to get around the festival if they take only Main Street and State Street?
Answer:

The leg opposite the 50° angle represents the distance along Main Street, the leg adjacent to the 50° angle represents the distance along State Street, and the hypotenuse of the triangle represents the 0.8 mile distance along Laura Street.

Using the ratio \(\frac{\text { opp }}{\text { hyp }}\) for 50°:
0.7660 = \(\frac{\text { opp }}{0.8}\)
0.6128 = opp

Using the ratio \(\frac{\text { adj }}{\text { hyp }}\) for 50°:
0.6428 = \(\frac{\text { adj }}{0.8}\)
0. 51424 = adj
The total distance of the detour:
0.6128 + 0.51424 = 1.12704
The total distance to travel around the festival along State Street and Main Street is approximately 1.1 miles.

Question 7.
A cable anchors a utility pole to the ground as shown in the picture. The cable forms an angle of 70° with the ground. The distance from the base of the utility pole to the anchor point on the ground is 3.8 meters. Approximately how long is the support cable?

Answer:
The hypotenuse of the triangle represents the length of the support cable in the diagram, and the leg adjacent to the given 70° angIe represents the distance between the anchor point and the base of the utility pole. Using the value of the ratio \(\frac{\text { adj }}{\text { hyp }}\) for 70°:
0.3420 = \(\frac{3.8}{\mathrm{hyp}}\)
hyp = 0. 3420
hyp = 11. 1111
The length of the support cable is approximately 11.1 meters long.

Question 8.
Indy says that the ratio of \(\frac{\text { opp }}{\text { adj }}\) for an angle of 0° has a value of 0 because the opposite side of the triangle has a length of 0. What does she mean?
Answer:
Indy’s triangle is not actually a triangle since the opposite side does not have length, which means that it does not exist. As the degree measure of the angle considered gets closer to 0°, the value of the ratio gets closer to 0.

Eureka Math Geometry Module Lesson 25 Exit Ticket Answer Key

Question 1.
Use the chart from the Exploratory Challenge to approximate the unknown lengths y and z to one decimal place.

Answer:

Question 2.
Why can we use the chart from the Exploratory Challenge to approximate the unknown lengths?
Answer:
The triangle in Problem 1 is similar to a triangle in the chart from the Exploratory Challenge. Since the triangles are similar, the values of the \(\frac{\text { opp }}{\text { hyp }}\) and \(\frac{\text { adj }}{\text { hyp }}\) ratios in reference to the acute angle of 28° can be used in the equations needed to solve for unknown lengths.

Eureka Math Geometry Module Lesson 25 Exploratory Challenge Ticket Answer Key

Note: Angle measures are approximations.
For each triangle in your set, determine missing angle measurements and side lengths. Side lengths should be measured to one decimal place. Make sure that each of the \(\frac{\text { opp }}{\text { hyp }}\) and \(\frac{\text { adj }}{\text { hyp }}\) ratios are set up and missing values are calculated and rounded appropriately.

Answer:

Answer:

With a partner, discuss what you can conclude about each pair of triangles between the two sets.
Answer:
Each pair of triangles is similar by the AA criterion. The marked acute angle for each pair of corresponding triangles is the same. The values of the \(\frac{\text { opp }}{\text { hyp }}\) and \(\frac{\text { adj }}{\text { hyp }}\) ratios are the same.

Engage NY Eureka Math Geometry Module 2 Lesson 32 Answer Key

Eureka Math Geometry Module 2 Lesson 32 Example Answer Key

Example 1.
A surveyor needs to determine the distance between two points A and B that lie on opposite banks of a river. A point C is chosen 160 meters from point A, on the same side of the river as A. The measures of ∠BAC and ∠ACB are 41° and 55°, respectively. Approximate the distance from A to B to the nearest meter.

Answer:
→ What measurement can we add to the diagram based on the provided information?
The measurement of ∠B must be 84° by the triangle sum theorem.

→ Use the law of sines to set up all possible ratios applicable to the diagram.
\(\frac{\sin 41}{a}\) = \(\frac{\sin 84}{160}\) = \(\frac{\sin 55}{c}\)

→ Which ratios will be relevant to determining the distance from A to B?
\(\frac{\sin 84}{160}\) = \(\frac{\sin 55}{c}\)

→ Solve for c.
c = \(\frac{160 \sin 55}{\sin 84}\)
c = 132
The distance from A to B is 132 m.

Example 2.
Our friend the surveyor from Example 1 is doing some further work. He has already found the distance between points A and B (from Example 1). Now he wants to locate a point D that is equidistant from both A and B and on the same side of the river as A. He has his assistant mark the point D so that ∠ABD and ∠BAD both measure 75°. What is the distance between D and A to the nearest meter?

Answer:

→ What do you notice about ∆ ABD right away?
∆ ABD must be an isosceles triangle since it has two angles of equal measure.

→ We must keep this in mind going forward. Add all relevant labels to the diagram.

Students should add the distance of 132 m between A and B and add the label of a and b to the appropriate sides.

→ Set up an equation using the law of cosines. Remember, we are trying to find the distance between D and or, as we have labeled it, b.
b² = 132² + a² – 2(132)(a) cos 75

→ Recall that this is an isosceles triangle; we know that a = b. To reduce variables, we will substitute b for a.
Rewrite the equation, and solve for b.
Sample solution:
b² = 132² + (b)² – 2(132) (b) cos 75
b² = 132² + (b)² – 264(b) cos 75
0 = 132² – 264(b)cos 75
264(b) cos 75 = 132²
b = \(\frac{132^{2}}{264 \cos 75}\)
b ≈ 255 m

Eureka Math Geometry Module 2 Lesson 32 Opening Exercise Answer Key

a. Find the length of d and e.

Answer:

b. Find the lengths of x and y. How is this different from part (a)?

Answer:
Accept any reasonable answer explaining that the triangle is not a right triangle; therefore, the trigonometric ratios used in part (a) are not applicable here.

Eureka Math Geometry Module 2 Lesson 32 Exercise Answer Key

Exercise 1.
In ∆ ABC, m∠A 30, a = 12, and b = 10. Find sin∠B. Include a diagram in your answer.

Answer:
\(\frac{\sin 30}{12}=\frac{\sin \angle B}{10}\)
sin ∠B = \(\frac{5}{12}\)

Exercise 2.
A car is moving toward a tunnel carved out of the base of a hill. As the accompanying diagram shows, the top of the hill, H, is sighted from two locations, A and B. The distance between A and B is 250 ft. What is the height, h, of the hill to the nearest foot?

Answer:

Let x represent BH, in feet. Applying the law of sines,

\(\overline{B H}\) is the hypotenuse of a 45 – 45 – 90 triangle whose sides are in the ratio 1: 1: √2, or h: h: h√2.
h√2 = x
h√2 = \(\frac{125}{\sin 15}\)
h = \(\frac{125}{\sin 15 \cdot \sqrt{2}}\)
h ≈ 342
The height of the hill is approximately 324 feet.

Exercise 3.
Parallelogram ABCD has sides of lengths 44 mm and 26 mm, and one of the angles has a measure of 100°.
Approximate the length of diagonal \(\overline{A C}\) to the nearest millimeter.

Answer:
In parallelogram ABCD, m∠C = 100°; therefore, m∠D = 80°.
Let d represent the length of \(\overline{A C}\).
d² = 44² + 26² – 2(44) (26) cos 80
d = 47
The length of \(\overline{A C}\) is 47 millimeters.

Eureka Math Geometry Module 2 Lesson 32 Problem Set Answer Key

Question 1.
Given ∆ ABC, AB = 14, m∠A = 57. 2°, and m∠C = 78. 4°, calculate the measure of angle B to the nearest tenth of a degree, and use the law of sines to find the lengths of \(\overline{A C}\) and \(\overline{B C}\) to the nearest tenth.

Answer:
By the angle sum of a triangle, m∠B = 44.4°.

Calculate the area of ∆ ABC to the nearest square unit.
Answer:
Area = \(\frac{1}{2}\) bc sin A
Area = \(\frac{1}{2}\) (10)(14) sin 57.2
Area = 70 sin 57.2
Area ≈ 59

Question 2.
Given ∆ DEF, m∠F = 39°, and EF = 13, calculate the measure of ∠E, and use the law of sines to find the lengths of \(\overline{D F}\) and \(\overline{D E}\) to the nearest hundredth.

Answer:
By the angle sum of a triangle, m∠E = 55°.

Question 3.
Does the law of sines apply to a right triangle? Based on ∆ ABC, the following ratios were set up according to the law of sines.

Fill in the partially completed work below:

What conclusions can we draw?
Answer:
The law of sines does apply to a right triangle. We get the formulas that are equivalent to sin ∠A = \(\frac{\text { opp }}{\text { hyp }}\) and sin ∠B = \(\frac{\text { opp }}{\text { hyp }}\) where A and B are the measures of the acute angles of the right triangle.

Question 4.
Given quadrilateral GHKJ, m∠H = 50°, m∠HKG = 80°, mLKGJ = 50°, ∠j is a right angle, and GH = 9 in, use the law of sines to find the length of \(\overline{G K}\), and then find the lengths of \(\overline{G J}\) and \(\overline{J K}\) to the nearest tenth of an inch.

Answer:
By the angle sum of a triangle, m∠HGK = 50°; therefore,
∆ GHK is an isosceles triangle since its base ∠‘s have equal measure.
\(\frac{\sin 50}{h}=\frac{\sin 80}{9}\)
h = \(\frac{9 \sin 50}{\sin 80}\)
h ≈ 7.0
k = 7 cos 50 ≈ 4.5
g = 7 sin 50 ≈ 5.4

Question 5.
Given triangle LMN, LM = 10, LN = 15, and m∠L = 38°, use the law of cosines to find the length of \(\overline{M N}\) to the nearest tenth.

Answer:
l² = 10² + 15² – 2(10)(15) cos 38
l² = 100 + 225 – 300 cos 38
l² = 325 – 300 cos 38
l² = 325 – 300 cos 38
l = \(\sqrt{325-300 \cos 38}\)
l ≈ 9.4
MN = 9.4
The length of \(\overline{M N}\) is approximately 9.4 units.

Question 6.
Given triangle ABC, AC = 6, AB = 8, and m∠A = 78°, draw a diagram of triangle ABC, and use the law of cosines to find the length \(\overline{B C}\).
Answer:

a² = 6² + 8² – 2(6)(8)(cos 78)
a² = 36 + 64 – 96(cos 78)
a² = 100 – 96 cos 78
a = \(\sqrt{100-96 \cos 78}\)
a ≈ 8.9
The length of BC is approximately 8.9 units.

Calculate the area of triangle ABC.
Area = \(\frac{1}{2}\)bc(sin A)
Area = \(\frac{1}{2}\)(6)(8)(sin 78)
Area = 23. 5(sin 78)
Area ≈ 23.5
The area of triangle ABC is approximately 23.5 square units.

Eureka Math Geometry Module 2 Lesson 32 Exit Ticket Answer Key

Question 1.
Use the law of sines to find lengths b and c in the triangle below. Round answers to the nearest tenth as necessary.

Answer:

Question 2.
Given ∆ DEF, use the law of cosines to find the length of the side marked d to the nearest tenth.

Answer:
d² = 6² + 9² – 2(6)(9) (cos 65)
d² = 36 + 81 – 108(cos 65)
d² = 117 – 108 (cos 65)
d = \(\sqrt{117-108(\cos 65)}\)
d ≈ 8.4

Engage NY Eureka Math Geometry Module 2 Lesson 31 Answer Key

Eureka Math Geometry Module 2 Lesson 31 Example Answer Key

Example 1.
Find the area of ∆ GHI.

Answer:

Allow students the opportunity and the time to determine what they must find (the height) and how to locate it (one option is to drop an altitude from vertex H to side \(\overline{GI}\)). For students who are struggling, consider showing just the altitude and allowing them to label the newly divided segment lengths and the height.

→ How can the height be calculated?
By applying the Pythagorean theorem to both of the created right triangles to find x,
h² = 49 – x²                h² = 144 – (15 – x)²
49 – x² = 144 – (15 – x)²
49 – x² = 144- 225 + 30x – x²
130 = 30x
x = \(\frac{13}{3}\)
HJ = \(\frac{13}{3}\), IJ = \(\frac{32}{3}\)

The value of x can then be substituted into either of the expressions equivalent to h² to find h.
h² = 49 – \(\left(\frac{13}{3}\right)^{2}\)
h² = 49 – \(\frac{169}{9}\)
h = \(\frac{4 \sqrt{17}}{3}\)

→ What is the area of the triangle?
Area = \(\left(\frac{1}{2}\right)(15)\left(\frac{4 \sqrt{17}}{3}\right)\)
Area = 10√17

Example 2.
A farmer is planning how to divide his land for planting next year’s crops. A triangular plot of land is left with two known
side lengths measuring 500 m and 1,700 m.
What could the farmer do next in order to find the area of the plot?
Answer:
→With just two side lengths known of the plot of land, what are the farmer’s options to determine the area of his plot of land?
He can either measure the third side length, apply the Pythagorean theorem to find the height of the triangle, and then calculate the area, or he can find the measure of the included angle between the known side lengths and use trigonometry to express the height of the triangle and then determine the area of the triangle.

→ Suppose the included angle measure between the known side lengths is 30°. What is the area of the plot of land? Sketch a diagram of the plot of land.

Area = \(\frac{1}{2}\) (1700)(500) sin 30
Area = 212 500
The area of the plot of land is 212,500 square meters.

Eureka Math Geometry Module 2 Lesson 31 Opening Exercise Answer Key

Three triangles are presented below. Determine the areas for each triangle, if possible. If it is not possible to find the area with the provided information, describe what Is needed in order to determine the area.

Answer:
The area of ∆ ABC is \(\frac{1}{2}\)(5)(12), or 30 square units, and the area of ∆ DEF is \(\frac{1}{2}\) (8)(2o), or 80 square units. There is not enough information to find the height of ∆ GHI and, therefore, the area of the triangle.

Is there a way to find the missing information?
Answer:
Without further information, there is no way to calculate the area.

Eureka Math Geometry Module 2 Lesson 31 Exercise Answer Key

Exercise 1.
A real estate developer and her surveyor are searching for their next piece of land to build on. They each examine a plot of land in the shape of ∆ ABC. The real estate developer measures the length of \(\overline{A B}\) and \(\overline{A C}\) and finds them to both be approximately 4,000 feet, and the included angle has a measure of approximately 50°. The surveyor measures the length of \(\overline{A C}\) and \(\overline{B C}\) and finds the lengths to be approximately 4,000 feet and 3,400 feet, respectively, and measures the angle between the two sides to be approximately 65°.

a. Draw a diagram that models the situation, labeling all lengths and angle measures.
Answer:

b. The real estate developer and surveyor each calculate the area of the plot of land and both find roughly the same area. Show how each person calculated the area; round to the nearest hundred. Redraw the diagram with only the relevant labels for both the real estate agent and surveyor.
Answer:

c. What could possibly explain the difference between the real estate agent’s and surveyor’s calculated areas?
Answer:
The difference in the area of measurements can be accounted for by the approximations of the measurements taken instead of exact measurements.

Eureka Math Geometry Module 2 Lesson 31 Problem Set Answer Key

Find the area of each triangle. Round each answer to the nearest tenth.

Question 1.

Answer:
Area = \(\frac{1}{2}\) (12)(9)(sin 21)
Area = 54(sin 21) ≈ 19.4
The area of the triangle is approximately 19.4 square units.

Question 2.

Answer:
Area = \(\frac{1}{2}\) (2)(11)(sin 34)
Area = 11(sin 34) ≈ 6.2
The area of the triangle is approximately 6.2 square units.

Question 3.

Answer:
Area = \(\frac{1}{2}\)(8) (6\(\frac{1}{2}\)) (sin 55)
Area = 26(sin 55) ≈ 21.3
The area of the triangle is approximately 21.3 square units.

Question 4.

Answer:
The included angle is 60° by the angle sum oía triangle.
Area = \(\frac{1}{2}\) (12)(6 + 6√3) sin 60
Area = 6(6 + 6√3) \(\left(\frac{\sqrt{3}}{2}\right)\)
Area = (36 + 36√3) \(\left(\frac{\sqrt{3}}{2}\right)\)
Area = 18√3 + 18(3)
Area = 18√3 + 54 ≈ 85.2
The area of the triangle is approximately 85.2 square units.

Question 5.
In ∆ DEF, EF = 15, DF = 20, and m∠F = 63°. Determine the area of the triangle. Round to the nearest tenth.
Answer:

Area = \(\frac{1}{2}\) (20) (15) sin(63) ≈ 133.7
The area of ∆ DEF is 133.7 units².

Question 6.
A landscape designer is designing a flower garden for a triangular area that is bounded on two sides by the client’s house and driveway. The length of the edges of the garden along the house and driveway are 18 ft. and 8 ft., respectively, and the edges come together at an angle of 80°. Draw a diagram, and then find the area of the garden to the nearest square foot.
Answer:

The garden is in the shape of a triangle in which the lengths of two sides and the included angle have been provided.
Area(ABC) = \(\frac{1}{2}\) (8 ft.)(18 ft.) sin 80
Area(ABC) = (72 sin 80) ft²
Area(ABC) ≈ 71 ft²

Question 7.
A right rectangular pyramid has a square base with sides of length 5. Each lateral face of the pyramid is an isosceles triangle. The angle on each lateral face between the base of the triangle and the adjacent edge is 75°. Find the surface area of the pyramid to the nearest tenth.
Answer:

Using tangent, the altitude of the triangle to the base of length 5 is equal to 2.5 tan 75.
Using tangent, the altitude of the triangle to the base of length 5 is equal to 2.5 tan 75.
Area = \(\frac{1}{2}\) bh
Area = \(\frac{1}{2}\) (5)(2.5 sin 75)
Area = 6. 25(sin 75)

The total surface area of the pyramid is the sum of the four lateral faces and the area of the square base:
SA = 4(6. 25(sin 75)) + 5²
SA = 25 sin 75 + 25
SA ≈ 49.1
The surface area of the right rectangular pyramid is approximately 49.1 square units.

Question 8.
The Pentagon building in Washington, DC, is built in the shape of a regular pentagon. Each side of the pentagon measures 921 ft. in length. The building has a pentagonal courtyard with the same center. Each wall of the center courtyard has a length of 356 ft. What is the approximate area of the roof of the Pentagon building?
Answer:

Let A₁ represent the area within the outer perimeter of the Pentagon building in square feet.

The area within the outer perimeter of the Pentagon building is approximately 1,459, 379 ft².

The area of the roof of the Pentagon building is approximately 1, 241, 333 ft².

Question 9.
A regular hexagon is inscribed in a circle with a radius of 7. Find the perimeter and area of the hexagon.
Answer:

The regular hexagon can be divided into six equilateral triangular regions with each side of the triangles having a length of 7. To find the perimeter of the hexagon, solve the following:
6 7 = 42, so the perimeter of the hexagon is 42 units.
To find the area of one equilateral triangle:
Area = \(\frac{1}{2}\) (7)(7) sin 60
Area = \(\frac{49}{2}\left(\frac{\sqrt{3}}{2}\right)\)
Area = \(\frac{49 \sqrt{3}}{4}\)
The area of the hexagon is six times the area of the equilateral triangle.
Total Area = 6\(\left(\frac{49 \sqrt{3}}{4}\right)\)
Total Area = \(\frac{147 \sqrt{3}}{2}\) ≈ 127.3
The total area of the regular hexagon is approximately 127.3 square units.

Question 10.
In the figure below, ∠AEB is acute. Show that Area (∆ ABC) = AC · BE · sin ∠AEB.

Answer:

Let θ represent the degree measure of angle AEB, and let h represent the altitude of ∆ ABC (and ∆ ABE).
Area(∆ ABC) = \(\frac{1}{2}\) · AC · h
sin θ = \(\frac{h}{B E}\) which implies that h = BE · sin θ.
Therefore, by substitution:
Area(∆ ABC) = \(\frac{1}{2}\) AC · BE · sin ∠AEB.

Question 11.
Let ABCD be a quadrilateral. Let w be the measure of the acute angle formed by diagonals \(\overline{AC}\) and \(\overline{BD}\). Show that
Area(ABCD) = \(\frac{1}{2}\) AC · BD · sin w.
(Hint: Apply the result from Problem 10 to ∆ ABC and ∆ ACD.)

Answer:
Let the intersection of \(\overline{AC}\) and \(\overline{BD}\) be called point P.
Using the results from Problem 10, solve the following:
Area(∆ ABC) = \(\frac{1}{2}\)AC · BP · sin w and
Area(∆ ADC) = \(\frac{1}{2}\)AC · PD · sin w
Area(ABCD) = (\(\frac{1}{2}\)AC · BP · sin w) + (\(\frac{1}{2}\)Ac · PD sin w)      Area is additive.
Area(ABCD) = (\(\frac{1}{2}\)Ac. sin w) · (BP + PD)     Distributive property
Area(ABCD) = (\(\frac{1}{2}\)Ac. sin w) · (BD)     Distance is additive
And commutative addition gives us Area(ABCD) = \(\frac{1}{2}\) · AC · BD · sin w.

Eureka Math Geometry Module 2 Lesson 31 Exit Ticket Answer Key

Question 1.
Given two sides of the triangle shown, having lengths of 3 and 7 and their included angle of 49°, find the area of the triangle to the nearest tenth.

Answer:
Area = \(\frac{1}{2}\) (3)(7)(sin 49)
Area = 10. 5(sin 49) ≈ 7.9
The area of the triangle is approximately 7.9 square units.

Question 2.
In isosceles triangle PQR, the base QR = 11, and the base angles have measures of 71.45°. Find the area of ∆ PQR to the nearest tenth.

Answer:
Drawing an altitude from P to midpoint M on \(\overline{Q R}\) cuts the isosceles triangle into two right triangles with
QM = MR = 5.5. Using tangent, solve the following:
tan 71.45 = \(\frac{P M}{5.5}\)
PM = 5. 5(tan 71.45)
Area = \(\frac{1}{2}\) bh
Area = \(\frac{1}{2}\) (11)(5. 5(tan 71.45))
Area = 30. 25(tan 71.45) ≈ 90. 1
The area of the isosceles triangle is approximately 90.1 square units.

Engage NY Eureka Math Geometry Module 2 Lesson 30 Answer Key

Eureka Math Geometry Module 2 Lesson 30 Example Answer Key

Example 1.
a. What common right triangle was probably modeled in the construction of the triangle in Figure 2? Use sin 53° ≈ 0.8.

Answer:

Right triangle with side lengths 3, 4, and 5, since 0.8 = \(\frac{4}{5}\).
→ What common right triangle was probably modeled in the construction of the triangle in Figure 2?

Though it may not be immediately obvious to students, part (a) is the same type of question as they completed in Exercises 1 – 2. The difference is the visual appearance of the value of sin 53 in decimal form versus in fraction form. Allow students time to sort through what they must do to answer part (a). Offer guiding questions and comments as needed such as the following:

→ Revisit Exercises 1 – 2. What similarities and differences do you notice between Example 1, part (a), and Exercises 1 – 2?

→ What other representation of 0.8 may be useful in this question?

Alternatively, students should also see that the value of sin S3° can be thought of as \(\frac{\text { opp }}{\text { hyp }}=\frac{0.8}{1}\). We proceed to answer part (a) using this fraction. If students have responses to share, share them as a whole class, or proceed to share the following solution:

→ To determine the common right triangle that was probably modeled in the construction of the triangle in Figure 2, using the approximation sin 53° ≈ 0.8 means we are looking for a right triangle with side-length relationships that are well known.

→ Label the triangle with given acute angle measure of approximately 53° as is labeled in the following figure. The hypotenuse has length 1, the opposite side has length 0.8, and the side adjacent to the marked angle is labeled as x.

→ How can we determine the value of x?
We can apply the Pythagorean theorem.

→ Solve for x.
(0.8)² + x² = (1)²
0.64 + x² = 1
x² = 0.36
x = 0.6

→ The side lengths of this triangle are 0.6, 0.8, and 1. What well-known right triangle matches these lengths?
Even though the calculations to determine the lengths of the triangle have been made, determining that this triangle is a 3 – 4 – 5 triangle is still a jump. Allow time for students to persevere after the answer. Offer guiding questions and comments as needed such as the following:

→ Sketch a right triangle with side lengths 6, 8, and 10, and ask how that triangle is related to the one in the problem.

→ List other triangle side lengths that are in the same ratio as a 6 – 8 – 10 triangle.

Students should conclude part (a) with the understanding that a triangle with an acute angle measure of approximately 53° is a 3 – 4 – 5 triangle.

b. The actual angle between the base and lateral faces of the pyramid is actually closer to 52°. Considering the age of the pyramid, what could account for the difference between the angle measure in part (a) and the actual measure?
Answer:
The Great Pyramid is approximately 4,500 years old, and the weight of each block is several tons. It is conceivable that over time, the great weight of the blocks caused the pyramids to settle and shift the lateral faces enough so that the angle is closer to 52° than to 53°.

C. Why do you think the architects chose to use a 3 – 4 – 5 as a model for the triangle?
Answer:
Answers may vary. Perhaps they used it (1) because it is the right triangle with the shortest whole-number side lengths to satisfy the converse of the Pythagorean theorem and (2) because of the aesthetic it offers.

Example 2.
Show why tan θ = \(\frac{\sin \theta}{\cos \theta}\)

Answer:
Allow students time to work through the reasoning independently before guiding them through an explanation. To provide more support, consider having the diagram on the board and then writing the following to start students off:

tan θ = \(\frac{\sin \theta}{\cos \theta}\) because
sin θ = \(\frac{a}{c}\) and cos θ = \(\frac{b}{c}\).
Then,
tan θ = \(\frac{\sin \theta}{\cos \theta}\)
tan θ = \(\frac{a}{b}\), which is what we found earlier.

→ If you are given one of the values sin O, cos O, or tan 0, we can find the other two values using the identities
sin² θ + cos² θ = 1 and tan θ = \(\frac{\sin \theta}{\cos \theta}\) or by using the Pythagorean theorem.

Eureka Math Geometry Module 2 Lesson 30 Exercise Answer Key

Exercise 1.
In a right triangle with acute angle of measure θ, sin θ = \(\frac{1}{2}\). What is the value of cos θ? Draw a diagram as part of your response.
Answer:

cos θ = \(\frac{\sqrt{3}}{2}\)

Exercise 2.
In a right triangle with acute angle of measure θ, sin θ = \(\frac{7}{9}\) What is the value of tan θ? Draw a diagram as part of your response.
Answer:

tan θ = \(\frac{7}{4 \sqrt{2}}=\frac{7 \sqrt{2}}{8}\)

Exercise 3.
In a right triangle with acute angle of measure θ, sin θ = \(\frac{1}{2}\), use the Pythagorean identity to determine the value of cos θ.
Answer:
sin²θ + cos²θ = 1
\(\left(\frac{1}{2}\right)^{2}\) + cos²θ = 1
\(\frac{1}{4}\) + cos²θ = 1
cos²θ = \(\frac{3}{4}\)
cos θ = \(\frac{\sqrt{3}}{2}\)

Exercise 4.
Given a right triangle with acute angle of measure θ, sin θ = \(\frac{7}{9}\) use the Pythagorean identity to determine the value of tan θ.
Answer:

Eureka Math Geometry Module 2 Lesson 30 Problem Set Answer Key

Question 1.
If cos θ = \(\frac{4}{5}\), find sin θ and tan θ.
Answer:

Question 2.
If sin θ = \(\frac{44}{125}\), find cos θ and tan θ.
Answer:

Question 3.
If tan θ = 5, find sin θ and cos θ.
Answer:
tan θ = 5 = \(\frac{5}{1}\), so the legs of a right triangle can be considered to have lengths of 5 and 1. Using the Pythagorean
theorem:

5² + 1² = hyp²
26 = hyp²
√26 = hyp
sin θ = \(\frac{5}{\sqrt{26}}=\frac{5 \sqrt{26}}{26}\); cos θ = \(\frac{1}{\sqrt{26}}=\frac{\sqrt{26}}{26}\)

Question 4.
If sin θ = \(\frac{\sqrt{5}}{5}\), find cos θ and tan θ.
Answer:

Question 5.
Find the missing side lengths of the following triangle using sine, cosine, and/or tangent. Round your answer to four decimal places.

Answer:
\(\frac{x}{12}\) = tan 27
x = 12 tan 27 ≈ 6.1143
\(\frac{12}{y}\) = sin 63
y = \(\frac{12}{\sin 63}\) ≈ 13.4679

Question 6.
A surveying crew has two points A and B marked along a roadside at a distance of 400 yd. A third point C is marked at the back corner of a property along a perpendicular to the road at B. A straight path joining C to A forms a 28° angle with the road. Find the distance from the road to point C at the back of the property and the distance from A to C using sine, cosine, and/or tangent. Round your answer to three decimal places.

Answer:
tan 28 = \(\frac{B C}{400}\)
BC = 400(tan28)
BC ≈ 212. 684
The distance from the road to the back of the property is approximately 212.684 yd.
cos 28 = \(\frac{400}{A C}\)
AC = \(\frac{400}{\cos 28}\)
AC ≈ 453.028
The distance from point C to point A is approximately 453.028 yd.

Question 7.
The right triangle shown is taken from a slice of a right rectangular pyramid with a square base.

a. Find the height of the pyramid (to the nearest tenth).
Answer:
sin 66 = \(\frac{h}{9}\)
h = 9(sin 66)
h ≈ 8.2
The height of the pyramid is approximately 8.2 units.

b. Find the lengths of the sides of the base of the pyramid (to the nearest tenth).
Answer:
The lengths of the sides of the base of the pyramid are twice the length of the short leg of the right triangle shown.
cos 66 = \(\frac{n}{9}\)
n = 9(cos 66)
length = 2(9 cos 66)
length = 18 cos 66
length ≈ 7.3
The lengths of the sides of the base are approximately 7.3 units.

c. Find the lateral surface area of the right rectangular pyramid.
Answer:
The faces of the prism are congruent isosceles triangles having bases of 18 cos 66 and height of 9.
Area = \(\frac{1}{2}\) bh
Area = \(\frac{1}{2}\) (18 cos 66)(9)
Area = 81 cos 66
Area ≈ 32.9
The lateral surface area of the right rectangular pyramid is approximately 32.9 square units.

Question 8.
A machinist is fabricating a wedge in the shape of a right triangular prism. One acute angle of the right triangular base is 33°, and the opposite side is 6.5 cm. Find the length of the edges labeled l and m using sine, cosine, and/or tangent. Round your answer to the nearest thousandth of a centimeter.

Answer:
sin 33 = \(\frac{6.5}{l}\)
l = \(\frac{6.5}{\sin 33}\)
l ≈ 11. 935
Distance l is approximately 11.935 cm.

tan 33 = \(\frac{6.5}{m}\)
m = \(\frac{6.5}{\tan 33}\)
m ≈ 10.009
Distance m is approximately 10.009 cm.

Question 9.
Let sin θ = \(\frac{1}{m}\) where 1, m > 0. Express tan θ and cos θ in terms of l and m.
Answer:

Eureka Math Geometry Module 2 Lesson 30 Exit Ticket Answer Key

Question 1.
If sin β = \(\frac{4 \sqrt{29}}{29}\), use trigonometric identities to find cos β and tan β.
Answer:

Question 2.
Find the missing side lengths of the following triangle using sine, cosine, and/or tangent. Round your answer to four decimal places.

Answer:
cos 70 = \(\frac{3}{y}\)
y = \(\frac{3}{\cos 70}\) ≈ 8.7714
tan 70 = \(\frac{x}{3}\)
x = 3(tan 70) ≈ 8. 2424

Engage NY Eureka Math Geometry Module 2 Lesson 29 Answer Key

Eureka Math Geometry Module 2 Lesson 29 Example Answer Key

Example 1.
Scott, whose eye level is 1.5 m above the ground, stands 30 m from a tree. The angle of elevation of a bird at the top of the tree is 36°. How far above ground is the bird?

Answer:

→ With respect to your diagram, think of the measurement you are looking for.
In our diagram, we are looking for BC.
→ How will you find BC?
I can use the tangent to determine BC in meters.
tan 36 = \(\frac{B C}{A C}\)
tan 36 = \(\frac{B C}{30}\)
30 tan 36 = BC
BC ≈ 21.8

→ Have we found the height at which the bird is off the ground?
No. The full height must be measured from the ground, so the distance from the ground to Scott’s eye level must be added to BC.
The height of the bird off of the ground is 1.5 m + 21.8 m = 23.3 m.

→ So, between the provided measurements, including the angle of elevation, and the use of the tangent ratio, we were able to find the height of the bird.

Example 2
From an angle of depression of 40°, John watches his friend approach his building while standing on the rooftop. The rooftop is 16 m from the ground, and John’s eye level is at about 1.8 m from the rooftop. What is the distance between John’s friend and the building?

Answer:
Make sure to point out the angle of depression in the diagram below. Emphasize that the 40° angle of depression is the angle between the line of sight and the line horizontal (to the ground) from the eye.

→ Use the diagram to describe the distance we must determine.
We are going to find BC in meters.

→ How will we find BC?
We can use the tangent: tan 40 = \(\frac{A B}{B C}\)
tan 40 = \(\frac{A B}{B C}\)
tan 40 = \(\frac{17.8}{B C}\)
BC = \(\)
BC ≈ 21.2

→ Again, with the assistance of a few measurements, including the angle of depression, we were able to determine the distance between John’s friend and the building, which is 21.2 meters.

Eureka Math Geometry Module 2 Lesson 29 Opening Exercise Answer Key

a. Use a calculator to find the tangent of θ. Enter the values, correct to four decimal places, in the last row of the table.

Answer:

Note to the teacher: Dividing the values in the first two rows provides a different answer than using the full decimal readout of the calculator.

b. The table from Lesson 29 is provided here for you. In the row labeled \(\frac{\sin \theta}{\cos \theta}\), divide the sine values by the cosine values. What do you notice?
Answer:
For each of the listed degree values, dividing the value of sin θ by the corresponding value of cos θ yields a value very close to the value of tan θ. The values divided to obtain \(\frac{\sin \theta}{\cos \theta}\) were approximations, so the actual
values might be exactly the same.

Eureka Math Geometry Module 2 Lesson 29 Exercise Answer Key

Exercise 1.
Standing on the gallery of a lighthouse (the deck at the top of a lighthouse), a person spots a ship at an angle of depression of 20°. The lighthouse is 28 m tall and sits on a cliff 45 m tall as measured from sea level. What is the horizontal distance between the lighthouse and the ship? Sketch a diagram to support your answer.

Answer:
Approximately 201 m

Exercise 2.
A line on the coordinate plane makes an angle of depression of 36°. Find the slope of the line correct to four decimal places.

Answer:
Choose a segment on the line, and construct legs of a right triangle parallel to the x- and y-axes, as shown. If m is the length of the vertical leg and n is the length of the horizontal leg, then tan 36 = \(\frac{m}{n}\). The line decreases to the right, so the value of the slope must be negative. Therefore,
slope = –\(\frac{m}{n}\) = – tan 36 ≈ – 0.7265.

Eureka Math Geometry Module 2 Lesson 29 Problem Set Answer Key

Question 1.
A line in the coordinate plane has an angle of elevation of 53°. Find the slope of the line correct to four decimal places.
Answer:

Since parallel lines have the same slope, we can consider the line that passes through the origin with an angle of inclination of 53°. Draw a vertex at (5, 0) on the x-axis, and draw a segment from (5, 0) parallel to the y-axis to the intersection with the line to form a right triangle. If the base of the right triangle is 5-units long, let the height of the triangle be represented by y.
tan 53 = \(\frac{x}{5}\)
5(tan 53) = x
The slope of the line is

slope = tan 53 ≈ 1.3270.

Question 2.
A line in the coordinate plane has an angle of depression of 25°. Find the slope of the line correct to four decimal places.
Answer:

A line crosses the x-axis, decreasing to the right with an angle of depression of 25°. Using a similar method as in Problem 1, a right triangle is formed. If the leg along the x-axis represents the base of the triangle, let h represent the height of the triangle.
Using tangent:
tan 25 = \(\frac{h}{5}\)
5(tan 25) = h
The slope of the line is negative since the line decreases to the right:

Question 3.
In Problems 1 and 2, why do the lengths of the legs of the right triangles formed not affect the slope of the line?
Answer:
When using the tangent ratio, the length of one leg of a right triangle can be determined in terms of the other leg. Let x represent the length of the horizontal leg of a slope triangle. The vertical leg of the triangle is then x tan θ, where θ is the measure of the angle of inclination or depression. The slope of the line is

Question 4.
Given the angles of depression below, determine the slope of the line with the Indicated angle correct to four decimal places.
a. 35° angle of depression
Answer:
tan 35 ≈ 0.7002
slope ≈ – 0.7002

b. 49° angle of depression
Answer:
tan 49 ≈ 1. 1504
slope ≈ – 1.1504

c. 80° angle of depression
Answer:
tan 80 ≈ 5. 6713
slope ≈ – 5. 6713

d. 87° angle of depression
Answer:
tan 87 ≈ 19. 0811
slope ≈ – 19. 0811

e. 89° angle of depression
Answer:
tan 89 ≈ 57. 2900
slope ≈ – 57.2900

f. 89.9° angle of depression
Answer:
tan 89.9 ≈ 572.9572
slope ≈ – 572.9572

g. What appears to be happening to the slopes (and tangent values) as the angles of depression get closer to 90°?
Answer:
As the angles get closer to 90°, their slopes (and tangent values) get much farther from zero.

h. Find the slopes of angles of depression that are even closer to 90° than 89.90. Can the value of the tangent of 90° be defined? Why or why not?
Answer:
Choices of angles will vary. The closer an angle is in measure to 90°, the greater the tangent value of that angle and the farther the slope of the line determined by that angle is from zero. An angle of depression of 90° would be a vertical line, and vertical lines have 0 run; therefore, the value of the ratio rise: run is undefined. The value of the tangent of 90° would have a similar outcome because the adjacent leg of the “triangle” would have a length of 0, so the ratio \(\frac{\text { opp }}{\text { adj }}=\frac{\text { opp }}{0}\), which is undefined.

Question 5.
For the indicated angle, express the quotient in terms of sine, cosine, or tangent. Then, write the quotient in simplest terms.

a. \(\frac{4}{2 \sqrt{13}}\); α
Answer:
cos α = \(\frac{4}{2 \sqrt{13}}=\frac{2}{\sqrt{13}}=\frac{2 \sqrt{13}}{13}\)

b. \(\frac{6}{4}\); α
Answer:
tan α = \(\frac{6}{4}=\frac{3}{2}\)

c. \(\frac{4}{2 \sqrt{13}}\); β
Answer:
sin β = \(\frac{4}{2 \sqrt{13}}=\frac{2 \sqrt{13}}{13}\)

d. \(\frac{4}{6}\); β
Answer:
tan β = \(\frac{4}{6}=\frac{2}{3}\)

Question 6.
The pitch of a roof on a home Is expressed as a ratio of vertical rise: horizontal run where the run has a length of 12 units. If a given roof design includes an angle of elevation of 22. 5° and the roof spans 36 ft. as shown in the diagram, determine the pitch of the roof. Then, determine the distance along one of the two sloped surfaces of the roof.

Answer:

The diagram as shown is an isosceles triangle since the base angles have equal measure. The altitude, a, of the triangle is the vertical rise of the roof.

The right triangles formed by drawing the altitude of the given isosceles triangle have a leg of length 18 ft.
tan 225 = \(\frac{a}{18}\)
a = 18 tan 22.5
a ≈ 7.5

Roof pitch:
\(\frac{7.5}{18}=\frac{h}{12}\)
18h = 12 · 7.5
h = 5

cos 22.5 = \(\frac{18}{s}\)
s = \(\frac{18}{\cos 22.5}\)
s ≈ 19.5

The pitch of the roof is 5:12.
The sloped surface of the roof has a distance of approximately 19.5 ft.

Question 7.
An anchor cable supports a vertical utility pole forming a 51° angle with the ground. The cable is attached to the top of the pole. If the distance from the base of the pole to the base of the cable is 5 meters, how tall is the pole?
Answer:
Let h represent the height of the pole in meters.
Using tangent:
tan 51 = \(\frac{h}{5}\)
5(tan 51) = h
6.17 ≈ h
The height of the utility pole is approximately 6. 17 meters.

Question 8.
A winch is a tool that rotates a cylinder, around which a cable is wound. When the winch rotates in one direction, it draws the cable in. Joey is using a winch and a pulley (as shown in the diagram) to raise a heavy box off the floor and onto a cart. The box is 2 ft. tall, and the winch is 14 ft. horizontally from where cable drops down vertically from the pulley. The angle of elevation to the pulley is 42°. What is the approximate length of cable required to connect the winch and the box?

Answer:
Let h represent the length of cable in the distance from the winch to the pulley along the hypotenuse of the right triangle shown in feet, and let y represent the distance from the pulley to the floor in feet.
Using tangent:
tan 42 = \(\frac{y}{14}\)
14(tan 42) = y
12.61 ≈ y

Using cosine:
cos 42 = \(\frac{14}{h}\)
h = \(\frac{14}{\cos 42}\)
h ≈ 18.84

Let t represent the total amount of cable from the winch to the box in feet:
t ≈ 12.61 + 18.84 – 2
t ≈ 29.45
The total length of cable from the winch to the box is approximately 29.45 ft.

Eureka Math Geometry Module 2 Lesson 29 Exit Ticket Answer Key

Question 1.
A line on the coordinate plane makes an angle of depression of 24°. Find the slope of the line correct to four decimal places.

Answer:
Choose a segment on the line, and construct legs of a right triangle parallel to the x- and y-axes, as shown. If m is the length of the vertical leg and n is the length of the horizontal leg, then tan 24 = \(\frac{m}{n}\) The line decreases to the right, so the value of the slope must be negative. Therefore,
slope = – \(\frac{m}{n}\) = – tan 24 ≈ – 0.4452.

Question 2.
Samuel is at the top of a tower and will ride a trolley down a zip line to a lower tower. The total vertical drop of the zip line is 40 ft. The zip line’s angle of elevation from the lower tower is 11. 5°. What is the horizontal distance between the towers?

Answer:

A right triangle is formed by the zip line’s path, the vertical drop along the upper tower, and the horizontal distance between the towers. Let x represent the horizontal distance between the towers in feet. Using tangent:
tan 11.5 = \(\frac{40}{x}\)
x = \(\frac{40}{\tan 11.5}\)
x ≈ 196.6
The horizontal distance between the towers ¡s approximately 196.6 ft.