Eureka Math Geometry Module 2 – Page 2

Eureka Math Geometry Module 2 Lesson 25 Answer Key

June 11, 2021

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.

$Eureka Math Geometry 2 Module 2 Lesson 25 Exercise Answer Key 1$

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.
$Eureka Math Geometry 2 Module 2 Lesson 25 Exercise Answer Key 2$
Answer:
$Eureka Math Geometry 2 Module 2 Lesson 25 Exercise Answer Key 3$

Exercise 5.
$Eureka Math Geometry 2 Module 2 Lesson 25 Exercise Answer Key 4$
Answer:
$Eureka Math Geometry 2 Module 2 Lesson 25 Exercise Answer Key 5$

Exercise 6.
$Eureka Math Geometry 2 Module 2 Lesson 25 Exercise Answer Key 6$
Answer:
$Eureka Math Geometry 2 Module 2 Lesson 25 Exercise Answer Key 7$

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.
$Eureka Math Geometry 2 Module 2 Lesson 25 Exercise Answer Key 8$
Answer:
$Eureka Math Geometry 2 Module 2 Lesson 25 Exercise Answer Key 9$

Exercise 8.
$Eureka Math Geometry 2 Module 2 Lesson 25 Exercise Answer Key 10$
Answer:
$Eureka Math Geometry 2 Module 2 Lesson 25 Exercise Answer Key 11$

Exercise 9.
$Eureka Math Geometry 2 Module 2 Lesson 25 Exercise Answer Key 12$
Answer:
$Eureka Math Geometry 2 Module 2 Lesson 25 Exercise Answer Key 13$

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.
$Eureka Math Geometry 2 Module 2 Lesson 25 Exercise Answer Key 14$
Answer:
$Eureka Math Geometry 2 Module 2 Lesson 25 Exercise Answer Key 15$
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.

$Eureka Math Geometry 2 Module 2 Lesson 25 Problem Set Answer Key 16$ Answer:
$Eureka Math Geometry 2 Module 2 Lesson 25 Problem Set Answer Key 17$

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.
$Eureka Math Geometry 2 Module 2 Lesson 25 Problem Set Answer Key 18$
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.
$Eureka Math Geometry 2 Module 2 Lesson 25 Problem Set Answer Key 19$
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.
$Eureka Math Geometry 2 Module 2 Lesson 25 Problem Set Answer Key 20$
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.
$Eureka Math Geometry 2 Module 2 Lesson 25 Problem Set Answer Key 21$
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°.
$Eureka Math Geometry 2 Module 2 Lesson 25 Problem Set Answer Key 22$
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:
$Eureka Math Geometry 2 Module 2 Lesson 25 Problem Set Answer Key 23$

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?
$Eureka Math Geometry 2 Module 2 Lesson 25 Problem Set Answer Key 24$
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.
$Eureka Math Geometry 2 Module 2 Lesson 25 Exit Ticket Answer Key 25$
Answer:
$Eureka Math Geometry 2 Module 2 Lesson 25 Exit Ticket Answer Key 26$

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.

$Eureka Math Geometry 2 Module 2 Lesson 25 Exploratory Challenge Ticket Answer Key 27$
Answer:
$Eureka Math Geometry 2 Module 2 Lesson 25 Exploratory Challenge Ticket Answer Key 28$

$Eureka Math Geometry 2 Module 2 Lesson 25 Exploratory Challenge Ticket Answer Key 29$
Answer:
$Eureka Math Geometry 2 Module 2 Lesson 25 Exploratory Challenge Ticket Answer Key 30$

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.

Eureka Math Geometry Module 2 Lesson 32 Answer Key

June 11, 2021

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.
$Eureka Math Geometry Module 2 Lesson 32 Example Answer Key 1$
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?
$Eureka Math Geometry Module 2 Lesson 32 Example Answer Key 2$
Answer:
$Eureka Math Geometry Module 2 Lesson 32 Example Answer Key 3$
→ 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.
$Eureka Math Geometry Module 2 Lesson 32 Opening Exercise Answer Key 4$
Answer:
$Eureka Math Geometry Module 2 Lesson 32 Opening Exercise Answer Key 5$

b. Find the lengths of x and y. How is this different from part (a)?
$Eureka Math Geometry Module 2 Lesson 32 Opening Exercise Answer Key 6$
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.
$Eureka Math Geometry Module 2 Lesson 32 Exercise Answer Key 7$
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?
$Eureka Math Geometry Module 2 Lesson 32 Exercise Answer Key 8$
Answer:
$Eureka Math Geometry Module 2 Lesson 32 Exercise Answer Key 9$
Let x represent BH, in feet. Applying the law of sines,
$Eureka Math Geometry Module 2 Lesson 32 Exercise Answer Key 10$
$\overline{B H}$ 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.
$Eureka Math Geometry Module 2 Lesson 32 Exercise Answer Key 11$
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.
$Eureka Math Geometry Module 2 Lesson 32 Problem Set Answer Key 12$
Answer:
By the angle sum of a triangle, m∠B = 44.4°.
$Eureka Math Geometry Module 2 Lesson 32 Problem Set Answer Key 13$

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.
$Eureka Math Geometry Module 2 Lesson 32 Problem Set Answer Key 14$
Answer:
By the angle sum of a triangle, m∠E = 55°.
$Eureka Math Geometry Module 2 Lesson 32 Problem Set Answer Key 15$

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.
$Eureka Math Geometry Module 2 Lesson 32 Problem Set Answer Key 16$
Fill in the partially completed work below:
$Eureka Math Geometry Module 2 Lesson 32 Problem Set Answer Key 17$
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.
$Eureka Math Geometry Module 2 Lesson 32 Problem Set Answer Key 18$
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.
$Eureka Math Geometry Module 2 Lesson 32 Problem Set Answer Key 19$
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:
$Eureka Math Geometry Module 2 Lesson 32 Problem Set Answer Key 20$
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.
$Eureka Math Geometry Module 2 Lesson 32 Exit Ticket Answer Key 21$
Answer:
$Eureka Math Geometry Module 2 Lesson 32 Exit Ticket Answer Key 22$

Question 2.
Given ∆ DEF, use the law of cosines to find the length of the side marked d to the nearest tenth.
$Eureka Math Geometry Module 2 Lesson 32 Exit Ticket Answer Key 23$
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

Eureka Math Geometry Module 2 Lesson 31 Answer Key

June 11, 2021

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.
$Eureka Math Geometry Module 2 Lesson 31 Example Answer Key 1$
Answer:
$Eureka Math Geometry Module 2 Lesson 31 Example Answer Key 2$

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.
$Eureka Math Geometry Module 2 Lesson 31 Example Answer Key 3$
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.
$Eureka Math Geometry Module 2 Lesson 31 Example Answer Key 4$
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:
$Eureka Math Geometry Module 2 Lesson 31 Exercise Answer Key 5$

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:
$Eureka Math Geometry Module 2 Lesson 31 Exercise Answer Key 6$

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.
$Eureka Math Geometry Module 2 Lesson 31 Problem Set Answer Key 7$
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.
$Eureka Math Geometry Module 2 Lesson 31 Problem Set Answer Key 8$
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.
$Eureka Math Geometry Module 2 Lesson 31 Problem Set Answer Key 9$
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.
$Eureka Math Geometry Module 2 Lesson 31 Problem Set Answer Key 10$
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:
$Eureka Math Geometry Module 2 Lesson 31 Problem Set Answer Key 11$
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:
$Eureka Math Geometry Module 2 Lesson 31 Problem Set Answer Key 12$
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:
$Eureka Math Geometry Module 2 Lesson 31 Problem Set Answer Key 13$
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:
$Eureka Math Geometry Module 2 Lesson 31 Problem Set Answer Key 14$
Let A₁ represent the area within the outer perimeter of the Pentagon building in square feet.
$Eureka Math Geometry Module 2 Lesson 31 Problem Set Answer Key 15$
The area within the outer perimeter of the Pentagon building is approximately 1,459, 379 ft².
$Eureka Math Geometry Module 2 Lesson 31 Problem Set Answer Key 16$
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:
$Eureka Math Geometry Module 2 Lesson 31 Problem Set Answer Key 17$
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.
$Eureka Math Geometry Module 2 Lesson 31 Problem Set Answer Key 18$
Answer:
$Eureka Math Geometry Module 2 Lesson 31 Problem Set Answer Key 19$
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.)
$Eureka Math Geometry Module 2 Lesson 31 Problem Set Answer Key 20$
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.
$Eureka Math Geometry Module 2 Lesson 31 Exit Ticket Answer Key 21$
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.
$Eureka Math Geometry Module 2 Lesson 31 Exit Ticket Answer Key 22$
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.

Eureka Math Geometry Module 2 Lesson 30 Answer Key

June 11, 2021

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.
$Eureka Math Geometry Module 2 Lesson 30 Example Answer Key 1$
Answer:
$Eureka Math Geometry Module 2 Lesson 30 Example Answer Key 2$
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.
$Eureka Math Geometry Module 2 Lesson 30 Example Answer Key 3$

→ 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 = $\frac{\sin \theta}{\cos \theta}$
$Eureka Math Geometry Module 2 Lesson 30 Example Answer Key 4$
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:
$Eureka Math Geometry Module 2 Lesson 30 Example Answer Key 5$

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:
$Eureka Math Geometry Module 2 Lesson 30 Exercise Answer Key 6$
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:
$Eureka Math Geometry Module 2 Lesson 30 Exercise Answer Key 6$
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:
$Eureka Math Geometry Module 2 Lesson 30 Exercise Answer Key 7$

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:
$Eureka Math Geometry Module 2 Lesson 30 Problem Set Answer Key 8$

Question 2.
If sin θ = $\frac{44}{125}$, find cos θ and tan θ.
Answer:
$Eureka Math Geometry Module 2 Lesson 30 Problem Set Answer Key 9$

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:
$Eureka Math Geometry Module 2 Lesson 30 Problem Set Answer Key 10$
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:
$Eureka Math Geometry Module 2 Lesson 30 Problem Set Answer Key 11$

Question 5.
Find the missing side lengths of the following triangle using sine, cosine, and/or tangent. Round your answer to four decimal places.
$Eureka Math Geometry Module 2 Lesson 30 Problem Set Answer Key 12$
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.
$Eureka Math Geometry Module 2 Lesson 30 Problem Set Answer Key 13$
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.
$Eureka Math Geometry Module 2 Lesson 30 Problem Set Answer Key 14$
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.
$Eureka Math Geometry Module 2 Lesson 30 Problem Set Answer Key 15$
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:
$Eureka Math Geometry Module 2 Lesson 30 Problem Set Answer Key 16$

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:
$Eureka Math Geometry Module 2 Lesson 30 Exit Ticket Answer Key 17$

Question 2.
Find the missing side lengths of the following triangle using sine, cosine, and/or tangent. Round your answer to four decimal places.
$Eureka Math Geometry Module 2 Lesson 30 Exit Ticket Answer Key 18$
Answer:
cos 70 = $\frac{3}{y}$
y = $\frac{3}{\cos 70}$ ≈ 8.7714
tan 70 = $\frac{x}{3}$
x = 3(tan 70) ≈ 8. 2424

Eureka Math Geometry Module 2 Lesson 29 Answer Key

June 11, 2021

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?
$Eureka Math Geometry Module 2 Lesson 29 Example Answer Key 1$
Answer:
$Eureka Math Geometry Module 2 Lesson 29 Example Answer Key 2$
→ 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?
$Eureka Math Geometry Module 2 Lesson 29 Example Answer Key 3$
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.
$Eureka Math Geometry Module 2 Lesson 29 Example Answer Key 4$
→ 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.
$Eureka Math Geometry Module 2 Lesson 29 Opening Exercise Answer Key 5$
Answer:
$Eureka Math Geometry Module 2 Lesson 29 Opening Exercise Answer Key 6$

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.
$Eureka Math Geometry Module 2 Lesson 29 Exercise Answer Key 7$
Answer:
Approximately 201 m
$Eureka Math Geometry Module 2 Lesson 29 Exercise Answer Key 8$

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.
$Eureka Math Geometry Module 2 Lesson 29 Exercise Answer Key 9$
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:
$Eureka Math Geometry Module 2 Lesson 29 Problem Set Answer Key 10$
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
$Eureka Math Geometry Module 2 Lesson 29 Problem Set Answer Key 11$
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:
$Eureka Math Geometry Module 2 Lesson 29 Problem Set Answer Key 12$
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:
$Eureka Math Geometry Module 2 Lesson 29 Problem Set Answer Key 13$

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
$Eureka Math Geometry Module 2 Lesson 29 Problem Set Answer Key 14$

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.
$Eureka Math Geometry Module 2 Lesson 29 Problem Set Answer Key 15$
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.
$Eureka Math Geometry Module 2 Lesson 29 Problem Set Answer Key 16$
Answer:
$Eureka Math Geometry Module 2 Lesson 29 Problem Set Answer Key 17$
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?
$Eureka Math Geometry Module 2 Lesson 29 Problem Set Answer Key 18$
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.
$Eureka Math Geometry Module 2 Lesson 29 Exit Ticket Answer Key 19$
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.
$Eureka Math Geometry Module 2 Lesson 29 Exit Ticket Answer Key 20$

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?
$Eureka Math Geometry Module 2 Lesson 29 Exit Ticket Answer Key 21$
Answer:
$Eureka Math Geometry Module 2 Lesson 29 Exit Ticket Answer Key 22$
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.

Eureka Math Geometry Module 2 End of Module Assessment Answer Key

June 11, 2021

Engage NY Eureka Math Geometry Module 2 End of Module Assessment Answer Key

Eureka Math Geometry Module 2 End of Module Assessment Answer Key

Question 1.
In the figure below, rotate ∆EAB about E by 180° to get ∆E$A^{\prime} B^{\prime}$. If $\overline{A^{\prime} B^{\prime}} \| \overline{C D}$, prove that ∆EAB – ∆EDC.

$Eureka Math Geometry Module 2 End of Module Assessment Answer Key 1$

Answer:
Triangles EAB and E$A^{\prime} B^{\prime}$. are congruent since a 180° rotation is a rigid motion. Since $\overline{A^{\prime} B^{\prime}} \| \overline{C D}$, m∠E$B^{\prime} A^{\prime}$ = m∠ECD, and m∠$B^{\prime}$E$A^{\prime}$ = m∠CED. So, △E$A^{\prime} B^{\prime}$ ~ △EDC by AA similarity criteria, and △EAB ~ △EDC by the transitive property of similarity

Question 2.
Answer the following questions based on the diagram below.

$Eureka Math Geometry Module 2 End of Module Assessment Answer Key 2$

a. Find the sine and cosine values of angles r and s. Leave the answers as fractions.
sin r°=
sin s°=
cos r°=
cos s°=
tan r°=
tan s°=
Answer:
sin r°= $\frac{15}{17}$
sin s°= $\frac{8}{17}$
cos r°= $\frac{8}{17}$
cos s°= $\frac{15}{17}$
tan r°= $\frac{15}{8}$
tan s°= $\frac{8}{15}$

b. Why is the sine of an acute angle the same value as the cosine of its complement?
Answer:
By definition, sine is the ratio of the opposite side: hypotenuse and cosine is the ratio of the adjacent side: hypotenuse; since the opposite side of an angle is the adjacent side of its complement, sinθ = cos(90 – θ).

c. Determine the measures of the angles to the nearest tenth of a degree in the right triangles below.

i. Determine the measure of ∠a.

$of Module Assessment Answer Key 3$

Answer:
m∠a ≈ 45.6°

ii. Determine the measure of ∠b.

$Eureka Math Geometry Module 2 End of Module Assessment Answer Key 4$

Answer:
m∠b ≈ 72.2°

iii. Explain how you were able to determine the measure of the unknown angle in part (i) or part (ii).
Answer:
For part (i), students should state that they had to use arccos to determine the unknown angle because the information given about the side lengths included the side adjacent to the unknown angle and the hypotenuse.

For part (ii), students should state that they had to use arcsin to determine the unknown angle because the information given about the side lengths included the side opposite to the unknown angle and the hypotenuse.

d. A ball is dropped from the top of a 45 ft. building. Once the ball is released, a strong gust of wind blew the ball off course, and it dropped 4 ft. from the base of the building.

i. Sketch a diagram of the situation.
Answer:
Sample sketch shown below.

$Eureka Math Geometry Module 2 End of Module Assessment Answer Key 7$

ii. By approximately how many degrees was the ball blown off course? Round your answer to the nearest whole degree.
Answer:
The wind blew the ball about 5˚ off course.

Question 3.
A radio tower is anchored by long cables called guy wires, such as $\overline{A B}$ in the figure below. Point A is 250 m from the base of the tower, and m∠BAC = 59°.

$Eureka Math Geometry Module 2 End of Module Assessment Answer Key 5$

a. How long is the guy wire? Round to the nearest tenth.
Answer:
cos 59 = $\frac{250}{A B}$
AB = $\frac{250}{\cos 59}$
AB ≈ 485.4
The guy wire is approximately 485.4 meters.

b. How far above the ground is it fastened to the tower?
Answer:
tan 59 = $\frac{B C}{250}$
BC = 250 tan 59
BC ≈ 416.1
It is fastened approximately 416.1 meters above the ground.

c. How tall is the tower, $\overline{D C}$, if m∠DAC = 71°?
Answer:
tan 71 = $\frac{D C}{250}$
DC = 250 tan 71
DC ≈ 726.1
The tower is approximately 726.1 meters tall.

Question 4.
The following problem is modeled after a surveying question developed by a Chinese mathematician during the Tang dynasty in the seventh century C.E.

A building sits on the edge of a river. A man views the building from the opposite side of the river. He measures the angle of elevation with a handheld tool and finds the angle measure to be 45°. He moves 50 feet away from the river and remeasures the angle of elevation to be 30°.

What is the height of the building? From his original location, how far away is the viewer from the top of the building? Round to the nearest whole foot.
Answer:

$Eureka Math Geometry Module 2 End of Module Assessment Answer Key 8$

The angle of depression from the top of the building to the man’s original spot is also 45°, and the angle of depression to his final position is 60°, so the difference of the angles is 15°. Let d represent the distance from the man’s position at the edge of the river to the top of the building, and let h represent the height of the building in feet.

Using the law of sines:
$\frac{\sin 30}{d}=\frac{\sin 15}{50}$
d = $\frac{50 \sin 30}{\sin 15}$ ≈ 96.6
The distance d is approximately 96.6 feet.
By the Pythagorean theorem, the distance d = h√2.
50 $\frac{\sin 30}{\sin 15}$ = h√2
$\frac{25}{\sin 15(\sqrt{2})}$ = h
h ≈ 68.3
The distance from the viewer to the top of the building is approximately 97 ft., and the height of the building is approximately 68 ft.

Question 5.
Prove the Pythagorean theorem using similar triangles. Provide a well-labeled diagram to support your justification.
Answer:
A right triangle ABC has side lengths a, b, and c. An altitude is drawn from C to the opposite side, dividing c into lengths x and c – x. Since the altitude from C divides the triangle into smaller similar right triangles by the AA criterion, then:
$\frac{a}{x}=\frac{c}{a}$
a² = cx
$\frac{b}{c-x}=\frac{c}{b}$
b² = c(c – x)
a² + b² = cx + c(c – x) = c²
a² + b² = c²

Question 6.
In right triangle ABC with ∠B a right angle, a line segment $B^{\prime} C^{\prime}$ connects side $\overline{A B}$ with the hypotenuse so that ∠A$B^{\prime} C^{\prime}$ is a right angle as shown. Use facts about similar triangles to show why cos $C^{\prime}$ = cos C.

$Eureka Math Geometry Module 2 End of Module Assessment Answer Key 6$

Answer:
By the AA criterion, △ABC ~ △A$B^{\prime} C^{\prime}$. Let r be the scale factor of the similarity
transformation. Then, $B^{\prime} C^{\prime}$ = r . BC, and A$C^{\prime}$ = r . AC. Thus,
cos $C^{\prime}$ = $\frac{B^{\prime} C^{\prime}}{A C^{\prime}}=\frac{r \cdot B C}{r \cdot A C}=\frac{B C}{A C}$ = cos C
cos $C^{\prime}$ = cos C

Question 7.
Terry said, “I will define the zine of an angle x as follows. Build an isosceles triangle in which the sides of equal length meet at angle x. The zine of x will be the ratio of the length of the base of that triangle to the length of one of the equal sides.” Molly said, “Won’t the zine of x depend on how you build the isosceles triangle?”

a. What can Terry say to convince Molly that she need not worry about this? Explain your answer.
Answer:
Isosceles triangles with vertex angle x are similar to each other; therefore, the value of zine x, the ratio of the length of the base of that triangle to the length of one of the equal sides, is the same for all such triangles.

b. Describe a relationship between zine and sin.
Answer:
zine x = 2 sin $\frac{1}{2}$x

Eureka Math Geometry Module 2 Lesson 33 Answer Key

June 11, 2021

Engage NY Eureka Math Geometry Module 2 Lesson 33 Answer Key

Eureka Math Geometry Module 2 Lesson 33 Opening Exercise Answer Key

Opening Exercise:

For each triangle shown below, identify the method (Pythagorean theorem, law of sines, law of cosines) you would use to find each length x.

$Eureka Math Geometry Module 2 Lesson 33 Opening Exercise Answer Key 3$

$Eureka Math Geometry Module 2 Lesson 33 Opening Exercise Answer Key 4$

Answer:

$Eureka Math Geometry Module 2 Lesson 33 Opening Exercise Answer Key 1$ law of sines

$Eureka Math Geometry Module 2 Lesson 33 Opening Exercise Answer Key 2$

Eureka Math Geometry Module 2 Lesson 33 Example Answer Key

Example 1:

Find the missing length in ∆ABC.

$Eureka Math Geometry Module 2 Lesson 33 Example Answer Key 5$

Answer:
→ Which method should we use to find length AB in the triangle shown below? Explain.
Provide a minute for students to discuss in pairs.

Yes. The law of sines can be used because we are given information about two of the angles and one side. We can use the triangle sum theorem to find the measure of ∠C, and then we can use that information about the value of the pair of ratios $\frac{\sin A}{a}=\frac{\sin C}{c}$. Since the values are equivalent, we can solve to find the missing length.

→ Why can’t we use the Pythagorean theorem with this problem?
We can only use the Pythagorean theorem with right triangles. The triangle in this problem is not a right triangle.

→ Why can’t we use the law of cosines with this problem?
The law of cosines requires that we know the lengths of two sides of the triangle. We are only given information about the length of one side.

→ Write the equation that allows us to find the length of $\overline{A B}$.
Let x represent the length of $\overline{A B}$.

$\frac{\sin 75}{2.93}=\frac{\sin 23}{x}$
cx = $\frac{2.93 \sin 23}{\sin 75}$

→ We want to perform one calculation to determine the answer so that it is most accurate and rounding errors are avoided. In other words, we do not want to make approximations at each step. Perform the calculation, and round the length to the tenths place.

The length of $\overline{A B}$ is approximately 1.2.

Example 2:

Find the missing length in ∆ABC.

$Eureka Math Geometry Module 2 Lesson 33 Example Answer Key 6$

Answer:
→ Which method should we use to find side $\overline{A C}$ in the triangle shown below? Explain.
Provide a minute for students to discuss in pairs.

We do not have enough information to use the law of sines because we do not have enough information to write any of the ratios related to the law of sines. However, we can use
b² = a² + c² – 2ac cos B because we are given the lengths of sides a and c and we know the angle measure for ∠B.

→ Write the equation that can be used to find the length of $\overline{A C}$, and determine the length using one calculation. Round your answer to the tenths place.

x² = 5.81² + 5.95² – 2(5.81)(5.95)cos 16
x = $\sqrt{5.81^{2}+5.95^{2}-2(5.81)(5.95) \cos 16}$
x ≈ 1.6

Eureka Math Geometry Module 2 Lesson 33 Exercise Answer Key

Exercises 1 – 6:

Use the laws of sines and cosines to find all missing side lengths for each of the triangles in the exercises below. Round your answers to the tenths place.

Exercise 1.
Use the triangle to the right to complete this exercise.

a. Identify the method (Pythagorean theorem, law of sines, law of cosines) you would use to find each of the missing lengths of the triangle. Explain why the other methods cannot be used.

$Eureka Math Geometry Module 2 Lesson 33 Exercise Answer Key 7$

Answer:
Law of sines
The Pythagorean theorem requires a right angle, which is not applicable to this problem. The law of cosines requires information about two side lengths, which is not given. Applying the law of sines requires knowing the measure of two angles and the length of one side.

b. Find the lengths of $\overline{A C}$ and $\overline{A B}$.
Answer:
By the triangle sum theorem, m∠A = 54°
Let b represent the length of side $\overline{A C}$.
$\frac{\sin 54}{3.31}=\frac{\sin 74}{b}$
b = $\frac{3.31 \sin 74}{\sin 54}$
b ≈ 3.9

Let C represent the length of side $\overline{A B}$
$\frac{\sin 54}{3.31}=\frac{\sin 52}{c}$
c = $\frac{3.31 \sin 52}{\sin 54}$
c ≈ 3.2

Exercise 2.
Your school is challenging classes to compete in a triathlon. The race begins with a swim along the shore and then continues with a bike ride for 4 miles. School officials want the race to end at the place it began, so after the 4-mile bike ride, racers must turn 30° and run 3.5 miles directly back to the starting point. What is the total length of the race? Round your answer to the tenths place.

$Eureka Math Geometry Module 2 Lesson 33 Exercise Answer Key 8$

a. Identify the method (Pythagorean theorem, law of sines, law of cosines) you would use to find the total length of the race. Explain why the other methods cannot be used.
Answer:
Law of cosines
The Pythagorean theorem requires a right angle, which is not applicable to this problem because we do not know if we have a right triangle. The law of sines requires information about two angle measures, which is not given. Applying the law of cosines requires knowing the measure of two sides and the included angle measure.

b. Determine the total length of the race. Round your answer to the tenths place.
Answer:
Let a represent the length of the swim portion of the tri at halon.
a² = 4² + 3.5² – 2(4)(3.5)cos30
a = $\sqrt{4^{2}+3.5^{2}-2(4)(3.5) \cos 30}$
a = 2.000322148 …
a ≈ 2
Total length of the race: 4 + 3.5 + (2) ≈ 9.5
The total length of the race is approximately 9.5 miles.

Exercise 3.
Two lighthouses are 30 miles apart on each side of shorelines running north and south, as shown. Each lighthouse keeper spots a boat in the distance. One lighthouse keeper notes the location of the boat as 40° east of south, and the other lighthouse keeper marks the boat as 32° west of south. What is the distance from the boat to each of the lighthouses at the time it was spotted? Round your answers to the nearest mile.

$Eureka Math Geometry Module 2 Lesson 33 Exercise Answer Key 9$

Answer:
Let x be the distance from the southern lighthouse to the boat.
$\frac{\sin 72}{30}=\frac{\sin 50}{x}$
x = $\frac{30 \sin 50}{\sin 72}$
x = 24.16400382…
The southern lighthouse is approximately 24 mi. from the boat.

Let y be the distance from the northern lighthouse to the boat.
$\frac{\sin 72}{30}=\frac{\sin 50}{x}$
y = $\frac{30 \sin 58}{\sin 72}$
y = 26.750716 12…
The northern lighthouse is approximately 27 mi.from the boat.
Let a be the distance from the northern lighthouse to the boat.
a² = 30² + 24² – 2(30)(24) . cos58
a = $\sqrt{30^{2}+24^{2}-2(30)(24) \cdot \cos 58}$
a = 26.700491 75…
The northern lighthouse is approximately 27 mi. from the boat.

Exercise 4.
A pendulum 18 in. in length swings 72° from right to left. What is the difference between the highest and lowest point of the pendulum? Round your answer to the hundredths place, and explain how you found it.

$Eureka Math Geometry Module 2 Lesson 33 Exercise Answer Key 10$

Answer:

$Eureka Math Geometry Module 2 Lesson 33 Exercise Answer Key 11$

At the bottom of a swing, the pendulum is perpendicular to the ground, and it bisects the 72° angle; therefore, the pendulum currently forms an angle of 36° with the vertical. By the triangle sum theorem, the angle formed by the pendulum and the horizontal is 54°. The sin 54 will give us the length from the top of the pendulum to where the horizontal and vertical lines intersect, x, which happens to be the highest point of the pendulum.
sin 54 = $\frac{x}{18}$
18 sin 54 = x
14.5623059… = x
The lowest point would be when the pendulum is perpendicular to the ground, which would be exactly 18 in. Then, the difference between those two points is approximately 3.44 in.

Exercise 5.
What appears to be the minimum amount of information about a triangle that must be given in order to use the law of sines to find an unknown length?
Answer:
To use the law of sines, you must know the measures of two angles and at least the length of one side.

Exercise 6.
What appears to be the minimum amount of information about a triangle that must be given in order to use the law of cosines to find an unknown length?
Answer:
To use the law of cosines, you must know at least the lengths of two sides and the angle measure of the included angle.

Eureka Math Geometry Module 2 Lesson 33 Problem Set Answer Key

Question 1.
Given triangle EFG, FG = 15, angIe E has a measure of 38°, and angle F has a measure of 72°, find the measures of the remaining sides and angle to the nearest tenth. Justify your method.

$Eureka Math Geometry Module 2 Lesson 33 Problem Set Answer Key 12$

Answer:
Using the angle sum of a triangle, the remaining angle G has a measure of 70°. The given triangle provides two angles and one side opposite a given angle, so it is appropriate to apply the law of sines.

$\frac{\sin 38}{15}=\frac{\sin 72}{E G}=\frac{\sin 70}{E F}$
$\frac{\sin 38}{15}=\frac{\sin 70}{E F}$
EF = $\frac{15 \sin 70}{\sin 38}$
EF ≈ 22.9

$\frac{\sin 38}{15}=\frac{\sin 72}{E G}$
EG = $\frac{15 \sin 72}{\sin 38}$
EG ≈ 23.2

Question 2.
Given triangle ABC, angle A has a measure of 75°, AC = 15.2, and AB = 24, find BC to the nearest tenth. Justify your method.

$Eureka Math Geometry Module 2 Lesson 33 Problem Set Answer Key 13$

Answer:
The given information provides the lengths of the sides of the triangle and an included angle, so it is appropriate to use the law of cosines.

BC² = 24² + 15.2² – 2(24)(15.2)(cos 75)
BC² = 576 + 231.04 – 729.6 cos 75
BC² = 807.04 – 729.6 cos 75
BC = $\sqrt{807.04-729.6 \cos 75}$
BC ≈ 24.9

Question 3.
James flies his plane from point A at a bearing of 32° east of north, averaging a speed of 143 miles per hour for 3 hours, to get to an airfield at point B. He next flies 69° west of north at an average speed of 129 miles per hour for 4.5 hours to a different airfield at point C.

$Eureka Math Geometry Module 2 Lesson 33 Problem Set Answer Key 14$

a. Find the distance from A to B.
Answer:
distance = rate . time
d = 143 . 3
d = 429
The distance from A to B is 429 mi.

b. Find the distance from B to C.
Answer:
distance = rate . time
d = 129 . 4.5
d = 580.5
The distance from B to C is 580.5 mi.

c. Find the measure of angle ABC.
Answer:
All lines pointing to the north/south are parallel; therefore, by alternate interior ∠’s, the return path from B to A is 32° west of south. Using angles on a line, this mentioned angle, the measure of the angle formed by the path from B to C with north (69°) and m∠ABC sum to 180; thus, m∠ABC = 79°.

d. Find the distance from C to A.
Answer:

$Eureka Math Geometry Module 2 Lesson 33 Problem Set Answer Key 15$

The triangle shown provides two sides and the included angle, so using the law of cosines,
b² = a² + c² – 2ac(cos B)
b² = (580.5)² + (429)² – 2(580.5)(429)(cos 79)
b² = 336980.25 + 184041 – 498069(cos 79)
b² = 521021.25 – 498069(cos 79)
b = $\sqrt{521021.25-498069(\cos 79)}$
b ≈ 652.7
The distance from C to A is approximately 652.7 mi.

e. What length of time can James expect the return trip from C to A to take?
Answer:
distance = rate . time
652.7 = 129 t₁
5.1 ≈ t₁

distance = rate . time
652.7 = 143t₂
4.6 ≈ t₂
James can expect the trip from C to A to take between 4.6 and 5. 1 hours.

Question 4.
Mark is deciding on the best way to get from point A to point B as shown on the map of Crooked Creek to go fishing. He sees that if he stays on the north side of the creek, he would have to walk around a triangular piece of private property (bounded by $\overline{A C}$ and $\overline{B C}$).

His other option is to cross the creek at A and take a straight path to B, which he knows to be a distance of 1.2 mi. The second option requires crossing the water, which is too deep for his boots and very cold. Find the difference in distances to help Mark decide which path is his better choice.

$Eureka Math Geometry Module 2 Lesson 33 Problem Set Answer Key 16$

Answer:
$\overline{A B}$ is 4.82° north of east, and $\overline{A C}$ is 24.39° east of north. The directions north and east are perpendicular, so the angles at point A form a right angle. Therefore, m∠CAB = 60.79°.

By the angle sum of a triangle, m∠PCA = 103.68°.
∠PCA and ∠BCA are angles on a line with a sum of 180°, s0 m∠BCA = 76.32°.
Also, by the angle sum of a triangle,
m∠ABC = 42.89°.
Using the law of sines:
$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$
$\frac{\sin 60.79}{a}=\frac{\sin 42.89}{b}=\frac{\sin 76.32}{1.2}$
b ≈ 0.8406
a ≈ 1.0780

Let d represent the distance from point A to B through point C in miles.
d = AC + BC
d ≈ 0.8406 + 1.0780
d ≈ 1.9
The distance from point A to B through point C is approximately 1.9 mi. This distance is approximately 0.7 mi. longer than walking a straight line from A to B.

Question 5.
If you are given triangle ABC and the measures of two of its angles and two of its sides, would it be appropriate to apply the law of sines or the law of cosines to find the remaining side? Explain.
Answer:
Case 1:
Given two angles and two sides, one of the angles being an included angle, it is appropriate to use either the law of sines or the law of cosines.

$Eureka Math Geometry Module 2 Lesson 33 Problem Set Answer Key 17$

Case 2:
Given two angles and the sides opposite those angles, the law of sines can be applied as the remaining anglecan be calculated using the angle sum of a triangle. The law of cosines then can also be applied as the previous calculation provides an included angle.

$Eureka Math Geometry Module 2 Lesson 33 Problem Set Answer Key 18$

Eureka Math Geometry Module 2 Lesson 33 Exit Ticket Answer Key

Question 1.
Given triangle MLK, KL = 8, KM = 7, and m∠K = 75°, find the length of the unknown side to the nearest tenth.
Justify your method.

$Eureka Math Geometry Module 2 Lesson 33 Exit Ticket Answer Key 19$

Answer:
The triangle provides the lengths of two sides and their included angles,
so using the law of cosines:
k² = 8² + 7² – 2(8)(7)(cos 75)
k² = 64 + 49 – 112(cos 75)
k² = 113 – 112(cos 75)
k = $\sqrt{113-112(\cos 75)}$
k ≈ 9.2

Question 2.
Given triangle ABC, m∠A = 36°, ∠B = 79°, and AC = 9, find the lengths of the unknown sides to the nearest tenth.

$Eureka Math Geometry Module 2 Lesson 33 Exit Ticket Answer Key 20$

Answer:
By the angle sum of a triangle, m∠C = 65°.
Using the law of sines:
$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$
$\frac{\sin 36}{a}=\frac{\sin 79}{9}=\frac{\sin 65}{c}$
a = $\frac{9 \sin 36}{\sin 79}$
a ≈ 5.4

c = $\frac{9 \sin 65}{\sin 79}$
c ≈ 8.3

AB ≈ 8.3 and BC ≈ 5.4

Eureka Math Geometry Module 2 Lesson 34 Answer Key

June 11, 2021

Engage NY Eureka Math Geometry Module 2 Lesson 34 Answer Key

Eureka Math Geometry Module 2 Lesson 34 Opening Exercise Answer Key

Opening Exercise:

a. Dan was walking through a forest when he came upon a sizable tree. Dan estimated he was about 40 meters away from the tree when he measured the angle of elevation between the horizontal and the top of the tree to be 35 degrees. If Dan is about 2 meters tall, about how tall is the tree?

$Eureka Math Geometry Module 2 Lesson 34 Opening Exercise Answer Key 1$
Answer:
Let x represent the vertical distance from Dan’s eye level to the top of the tree.
tan 35 = $\frac{x}{40}$
40 tan 35 = x
28 ≈ x
The height of the tree is approximately 30 m.

b.
Dan was pretty impressed with this tree until he turned around and saw a bigger one, also 40 meters away in the other direction. “Wow,” he said. “I bet that tree is at least 50 meters tall!” Then, he thought a moment. “Hmm … if it is 50 meters tall, I wonder what angle of elevation I would measure from my eye level to the top of the tree?” What angle will Dan find if the tree is 50 meters tall? Explain your reasoning.

$Eureka Math Geometry Module 2 Lesson 34 Opening Exercise Answer Key 2$

Answer:
Let x represent the angle measure from the horizontal to the top of the tree.
tan x = $\frac{50}{40}$
tan x = $\frac{5}{4}$
tan x = 1.25

Eureka Math Geometry Module 2 Lesson 34 Exercise Answer Key

Exercises 1 – 5:

Exercise 1.
Find the measure of angles a through d to the nearest degree.

a. $Eureka Math Geometry Module 2 Lesson 34 Exercise Answer Key 3$
Answer:
arccos($\frac{13}{20}$) ≈ 49 m∠a

b. $Eureka Math Geometry Module 2 Lesson 34 Exercise Answer Key 4$
Answer:
arcsin ($\frac{40}{42}$) ≈ 72 m∠b

c. $Eureka Math Geometry Module 2 Lesson 34 Exercise Answer Key 5$
Answer:
arctan ($\frac{14}{29}$) ≈ 26° m∠c

d. $Eureka Math Geometry Module 2 Lesson 34 Exercise Answer Key 6$
Answer:
Several solutions are acceptable. One is shown below.
arccos ($\frac{51}{85}$) ≈ 53, m∠d = 53°

Exercise 2.
Shelves are being built in a classroom to hold textbooks and other supplies. The shelves will extend 10 in. from the wall. Support braces will need to be installed to secure the shelves. The braces will be attached to the end of the shelf and secured 6 in. below the shelf on the wall. What angle measure will the brace and the shelf make?

$Eureka Math Geometry Module 2 Lesson 34 Exercise Answer Key 7$

Answer:
arctan($\frac{6}{10}$) ≈ 31
The angle measure between the brace and the shelf is 31°.

Exercise 3.
A 16 ft. ladder leans against a wall. The foot of the ladder is 7 ft. from the wall.

a. Find the vertical distance from the ground to the point where the top of the ladder touches the wall.
Answer:
Let x represent the distance from the ground to the point where the top of the ladder touches the wall.
16² = 7² + x²
16² – 7² = x²
207 = x²
14 ≈ x
The top of the ladder is 14 ft. above the ground.

b. Determine the measure of the angle formed by the ladder and the ground.
Answer:
arccos ($\frac{7}{16}$) ≈ 64
The angle formed by the ladder and the ground is approximately 64°.

Exercise 4.
A group of friends have hiked to the top of the Mile High Mountain. When they look down, they can see their campsite, which they know is approximately 3 miles from the base of the mountain.

a. Sketch a drawing of the situation.
Answer:
$Eureka Math Geometry Module 2 Lesson 34 Exercise Answer Key 7$

b. What is the angle of depression?
Answer:
arctan ($\frac{3}{1}$) ≈ 72
The angle of depression is approximately 18°.

Exercise 5.
A roller coaster travels 80 ft. of track from the loading zone before reaching its peak. The horizontal distance between the loading zone and the base of the peak is 50 ft.

a. Model the situation using a right triangle.
Answer:

$Eureka Math Geometry Module 2 Lesson 34 Exercise Answer Key 8$

b. At what angle is the roller coaster rising according to the model?
Answer:
arccos ($\frac{50}{80}$) ≈ 51
The roller coaster is rising at approximately 51°.

Eureka Math Geometry Module 2 Lesson 34 Problem Set Answer Key

Question 1.
For each triangle shown, use the given information to find the indicated angle to the nearest degree.

a. $Eureka Math Geometry Module 2 Lesson 34 Problem Set Answer Key 10$
Answer:
tan θ = ($\frac{6}{7}$)
arctan ($\frac{6}{7}$) = θ
θ ≈ 41°

b. $Eureka Math Geometry Module 2 Problem Set Lesson 34 Problem Set Answer Key 25$
Answer:
cos θ = $\frac{3.6}{9.4}$
arccos(($\frac{6}{7}$) = θ
θ ≈ 67°

c. $Eureka Math Geometry Module 2 Lesson 34 Problem Set Answer Key 11$
Answer:
sin θ = $\frac{3}{11.4}$
arcsin ($\frac{3}{11.4}$) = θ
θ ≈ 15°

Question 2.
Solving a right triangle means using given information to find all the angles and side lengths of the triangle. Use arcsin and arccos, along with the given information, to solve right triangle ABC if leg AC = 12 and hypotenuse AB = 15.
Answer:

$Eureka Math Geometry Module 2 Lesson 34 Problem Set Answer Key 12$

By the Pythagorean theorem, BC = 9.
sin B = $\frac{12}{15}$
arcsin($\frac{12}{15}$) = m∠B
m∠B ≈ 53°

cos A = $\frac{12}{15}$
arccos($\frac{12}{15}$) = m∠A
m∠A ≈ 37°

b. Once you have found the measure of one of the acute angles in the right triangle, can you find the measure of the other acute angle using a different method than those used in this lesson? Explain.
Answer:
Yes. We could use the angle sum of a triangle after finding the measure of one acute angle.

Question 3.
A pendulum consists of a spherical weight suspended at the end of a string whose other end is anchored at a pivot point P. The distance from P to the center of the pendulum’s sphere, B, is 6 inches. The weight is held so that the string is taut and horizontal, as shown to the right, and then dropped.

$Eureka Math Geometry Module 2 Lesson 34 Problem Set Answer Key 13$

a. What type of path does the pendulum’s weight take as it swings?
Answer:
Since the string is a constant length, the path of the weight is circular.

b. Danni thinks that for every vertical drop of 1 inch that the pendulum’s weight makes, the degree of rotation is 15°. Do you agree or disagree with Danni? As part of your explanation, calculate the degree of rotation for every vertical drop of 1 inch from 1 inch to 6 inches.
Answer:
Disagree A right triangle can model the pendulum and its vertical drops as shown in the diagrams.

$Eureka Math Geometry Module 2 Lesson 34 Problem Set Answer Key 14$
The angle of rotation about P for a vertical drop of 1 inch is equal to the arcsin ($\frac{1}{6}$), which is approximately 10°.

$Eureka Math Geometry Module 2 Lesson 34 Problem Set Answer Key 15$
The angle of rotation about P for a vertical drop of 2 inches is equal to the arcsin ($\frac{2}{6}$), which is approximately 20°.

$Eureka Math Geometry Module 2 Lesson 34 Problem Set Answer Key 16$
The angle of rotation about P for a vertical drop of 3 inches is equal to the arcs in ($\frac{3}{6}$), which is 30°.

$Eureka Math Geometry Module 2 Lesson 34 Problem Set Answer Key 17$
The angle of rotation about P for a vertical drop of 4 inches is equal to the arcsin ($\frac{4}{6}$), which is approximately 42°.

$Eureka Math Geometry Module 2 Lesson 34 Problem Set Answer Key 18$
The angle of rotation about P for a vertical drop of 5 inches is equal to the arcsin ($\frac{5}{6}$), which is approximately 56°.

$Eureka Math Geometry Module 2 Lesson 34 Problem Set Answer Key 19$
The pendulum’s weight will be at its maximum distance below the pivot point, which means that the weight must be directly below the pivot point. This means that the string would be perpendicular to the horizontal starting position; therefore, the degree of rotation would be 90°.

Question 4.
A stone tower was built on unstable ground, and the soil beneath It settled under its weight, causing the tower to lean. The cylindrical tower has a diameter of 17 meters. The height of the tower on the low side measured 46. 3 meters and on the high side measured 47. 1 meters. To the nearest tenth of a degree, find the angle that the tower has leaned from its original vertical position.

$Eureka Math Geometry Module 2 Lesson 34 Problem Set Answer Key 20$

Answer:
The difference in, heights from one side of the tower to the other is 47.1 m – 46.3 m = 0.8m.
Model the difference in heights and the diameter of the tower using a right trlangla (The right triangle shown below is not drawn to scale).

$Eureka Math Geometry Module 2 Lesson 34 Problem Set Answer Key 21$

The unknown value θ represents the degree measure that the tower has leaned.
sin θ = $\frac{0.8}{17}$
θ = arcsin($\frac{0.8}{17}$)
θ ≈ 3
The tower has leaned approximately 3° from its vertical position.

Question 5.
Doug is installing a surveillance camera inside a convenience store. He mounts the camera 8 ft. above the ground and 15 ft. horizontally from the store’s entrance. The camera is meant to monitor every customer who enters and exits the store. At what angle of depression should Doug set the camera to capture the faces of all customers?
Note: This is a modeling problem and therefore will have various reasonable answers.
Answer:
The solution below represents only one of many possible reasonable solutions.
Most adults are between 4$\frac{1}{2}$ ft. and 6$\frac{1}{2}$ ft. tall, so the camera should be aimed to capture images within the range of 1$\frac{1}{2}$ ft. to 3$\frac{1}{2}$ ft. below its mounted height. Cameras capture a range of images, so Doug should mount the camera so that it points at a location in the doorway 2$\frac{1}{2}$ ft. below its mounted height.

$Eureka Math Geometry Module 2 Problem Set Answer Key 22$

The angle of depression is equal to the arctan $\left(\frac{2 \frac{1}{2}}{15}\right)$ which is approximately 10°.

Eureka Math Geometry Module 2 Lesson 34 Exit Ticket Answer Key

Question 1.
Explain the meaning of the statement “arcsin($\frac{1}{2}$) = 30°.” Draw a diagram to support your explanation.
Answer:

$Eureka Math Geometry Module 2 Exit Ticket Answer Key 23$

This means that the measure of the angle that has a sine ratio equal to $\frac{1}{2}$ is 30°.

Question 2.
Gwen has built and raised a wall of her new house. To keep the wall standing upright while she builds the next wall, she supports the wall with a brace, as shown in the diagram below. What is the value of p, the measure of the angle formed by the brace and the wall?

$Eureka Math Geometry Module 2 Exit Ticket Answer Key 24$

Answer:
arctan $\frac{5.5}{3}$ = p
61 ≈ p
The measure of the angle formed by the brace and the wall is approximately 61°.

Eureka Math Geometry Module 2 Mid Module Assessment Answer Key

June 11, 2021

Eureka Math Geometry Module 2 Mid Module Assessment Answer Key

Question 1.
The coordinates of ∆ABC are shown on the coordinate plane below. ∆ABC is dilated from the origin by scale factor r = 2.

$Eureka Math Geometry Module 2 Mid Module Assessment Answer Key 1$

a. Identify the coordinates of the dilated ∆$A^{\prime} B^{\prime} C^{\prime}$.
Answer:
Point A = (3, 2), then $A^{\prime}$ = (2×(3), 2×(2)) = (6, 4)
Point B = (0, -1), then $B^{\prime}$ = (2×(0), 2×(–1)) = (0, -2)
Point 𝐠= (-3, 1), then $C^{\prime}$ = (2×(–3), 2×(1)) = (-6, 2)

b. Is a ∆$A^{\prime} B^{\prime} C^{\prime}$ – ∆ABC? Explain.
Answer:
Yes. The side lengths of ∆$A^{\prime} B^{\prime} C^{\prime}$ are each two times the length of the sides of ∆ABC, and corresponding sides are proportional in length. Also, the corresponding angles are equal in measurement because dilations preserve the measurements of angles.

Question 2.
Points A, B, and C are not collinear, forming ∠BAC. Extend $\overrightarrow{A B}$ to point P. Line l passes through P and is parallel to segment BC. It meets $\overrightarrow{A B}$ at point Q.

a. Draw a diagram to represent the situation described.
Answer:

$Eureka Math Geometry Module 2 Mid Module Assessment Answer Key 12$

b. Is $\overline{P Q}$ longer or shorter than $\overline{B C}$?
Answer:

$\overline{P Q}$ is longer than$\overline{B C}$.

c. Prove that ∆ABC ~ ∆APQ.
Answer:
$\overline{P Q} \| \overline{B C}$. Since corresponding angles are equal in measure, then m∠ABC = m∠APQ. Additionally, m∠A = m∠A. Then ∆ABC ~ ∆APQ by AA similarity criterion.

d. What other pairs of segments in this figure have the same ratio of lengths that $\overline{P Q}$ has to $\overline{B C}$?
Answer:
PA : BA, QA : CA

Question 3.
There is a triangular floor space ∆ABC in a restaurant. Currently, a square portion DEFG is covered with tile. The owner wants to remove the existing tile and then tile the largest square possible within ABC, keeping one edge of the square on $\overline{A C}$.

a. Describe a construction that uses a dilation with center A that can be used to determine the maximum square $D^{\prime} E^{\prime} F^{\prime} G^{\prime}$ within ∆ABC with one edge on $\overline{A C}$.

$Eureka Math Geometry Module 2 Mid Module Assessment Answer Key 2$

Answer:

Use A as a center of dilation.
Draw $\overrightarrow{A F}$ through BC.
Label the intersection of $\overrightarrow{A F}$ and BC as $F^{\prime}$.
Construct $\overline{E^{\prime} F^{\prime}}$ parallel to EF, where $E^{\prime}$ is the intersection of AB and the parallel line.
Construct $\overline{F^{\prime} G^{\prime}}$ parallel to FG, where $G^{\prime}$ is the intersection of AC and the parallel line.
Construct $\overline{E^{\prime} D^{\prime}}$ parallel to ED, where $D^{\prime}$ is the intersection of AC and the parallel line.
Connect $D^{\prime}$, $E^{\prime}$, $F^{\prime}$, $G^{\prime}$.

b. What is the scale factor of $\overline{F G}$ to $\overline{F^{\prime} G^{\prime}}$ in terms of the distances $\overline{A F}$ and $\overline{A F^{\prime}}$?
Answer:
$\frac{A F^{\prime}}{A F}$

c. The owner uses the construction in part (a) to mark off where the square would be located. He measures AE to be 15 feet and E$E^{\prime}$ to be 5 feet. If the original square is 144 square feet, how many square feet of tile does he need for $D^{\prime} E^{\prime} F^{\prime} G^{\prime}$?
Answer:
The distance A$E^{\prime}$ = 20 ft, so the scale factor from DEFG to $D^{\prime} E^{\prime} F^{\prime} G^{\prime}$ is $\frac{20}{15}=\frac{4}{3}$.

Area ($D^{\prime} E^{\prime} F^{\prime} G^{\prime}$) = $\left(\frac{4}{3}\right)^{2}$ Area (DEFG)

Area ($D^{\prime} E^{\prime} F^{\prime} G^{\prime}$) = $\frac{16}{9}$(144)

Area ($D^{\prime} E^{\prime} F^{\prime} G^{\prime}$) = 256
The owner needs 256 square feet of tile for $D^{\prime} E^{\prime} F^{\prime} G^{\prime}$.

Question 4.
ABCD is a parallelogram, with the vertices listed counterclockwise around the figure. Points M, N, O, and P are the midpoints of sides $\overline{A B}$, $\overline{B C}$, $\overline{C D}$, and $\overline{D A}$, respectively. The segments MO and NP cut the parallelogram into four smaller parallelograms, with the point W in the center of ABCD as a common vertex.

$Eureka Math Geometry Module 2 Mid Module Assessment Answer Key 13$

a. Exhibit a sequence of similarity transformations that takes ∆AMW to ∆CDA. Be specific in describing the parameter of each transformation; for example, if describing a reflection, state the line of reflection.
Answer:
Answers will vary (e.g., 180˚ rotation about W; dilation with center of dilation C).

b. Given the correspondence in ∆AMW similar to ∆CDA, list all corresponding pairs of angles and corresponding pairs of sides. What is the ratio of the corresponding pairs of angles? What is the ratio of the corresponding pairs of sides?
Answer:
∠MAW corresponds to ∠DCA; ∠AMW corresponds to ∠CDA; ∠AWM corresponds to ∠CAD. $\overline{A M}$ corresponds to $\overline{C D}$; $\overline{M W}$ corresponds to $\overline{D A}$; $\overline{W A}$ corresponds to $\overline{A C}$.
The ratio of corresponding pairs of angles is 1 : 1.
The ratio of corresponding pairs of sides is 1 : 2.

Question 5.
Given two triangles, ∆ABC and ∆DEF, m∠CAB = m∠FDE, and m∠CBA = m∠FED. Points A, B, D, and E lie on line l as shown. Describe a sequence of rigid motions and/or dilations to show that ∆ABC ~ ∆DEF, and sketch an image of the triangles after each transformation.

$Eureka Math Geometry Module 2 Mid Module Assessment Answer Key 3$

Answer:
Reflect ΔDEF over the perpendicular bisector of $\overline{B E}$. The reflection takes E to B and D to a
point on line AB. Since angle measures are preserved in rigid motions, F must map to a point on $\overline{B C}$.

$Eureka Math Geometry Module 2 Mid Module Assessment Answer Key 14$

→ By the hypothesis, m∠A = m∠D; therefore, $\overline{D^{\prime} F^{\prime}} \| \overline{A C}$ since corresponding angles are equal in measure.
→ Since dilations map a segment to a parallel line segment, dilate ∆$D^{\prime} E^{\prime} F^{\prime}$ about $E^{\prime}$ and by scale factor r = $\frac{B A}{B D^{\prime}}$ and that sends $D^{\prime}$ to A.
→ By the dilation theorem, $F^{\prime}$ goes to C.

Question 6.
∆JKL is a right triangle; $\overline{N P}$ ⊥$\overline{K L}$, $\overline{N O}$ ⊥ $\overline{J K}$, $\overline{M N}$ || $\overline{O P}$

a. List all sets of similar triangles. Explain how you know.

$Eureka Math Geometry Module 2 Mid Module Assessment Answer Key 4$

Answer:
Set 1

∆MNO
∆PON
∆OPK

Set 2

∆JON
∆JKL
∆NPL
The triangles are similar because of the AA criterion.

b. Select any two similar triangles, and show why they are similar.
Answer:
$Eureka Math Geometry Module 2 Mid Module Assessment Answer Key 15$

Possible response:

∆MNO ~ ∆PON by the AA criterion.
∠K is a right angle since ∆JKL is a right triangle.
∠MON is a right angle since $\overline{\mathrm{NO}} \perp \overline{\mathrm{JK}}$
m∠NMO = m∠POK since $\overline{M N} \| \overline{O P}$̅, and $\overline{J K}$ is a transversal that intersects $\overline{M N}$ and $\overline{O P}$; corresponding ∠’s are equal in measure.
Therefore, by the AA criterion, ∆MNO ~ ∆PON.

Question 7.

a. The line PQ contains point O. What happens to $\overleftrightarrow{P Q}$ with a dilation about O and scale factor of r = 2? Explain your answer.

$Eureka Math Geometry Module 2 Mid Module Assessment Answer Key 5$

Answer:

Since the points P and Q are collinear with the center O, then by definition of a dilation, both P’ and Q’ will also be collinear with the center O.
The line PQ maps to itself.

b. The line PQ contains point O. What happens to $\overleftrightarrow{P Q}$ with a dilation about O and scale factor of r = 2? Explain your answer.

$Eureka Math Geometry Module 2 Mid Module Assessment Answer Key 6$

Answer:
The line PQ maps to a parallel line $P^{\prime} Q^{\prime}$.

Question 8.
Use the diagram below to answer the following questions.

a. State the pair of similar triangles. Which similarity criterion guarantees their similarity?
Answer:
∆DEF ~ ∆FGH
SAS

b. Calculate DE to the hundredths place.

$Eureka Math Geometry Module 2 Mid Module Assessment Answer Key 7$

Answer:
$\frac{D E}{F G}=\frac{D F}{F H}$
$\frac{D E}{6}=\frac{17}{8}$
8 DE = 102
DE ≈ 12.75

Question 9.
In ∆ABC, m∠A is 40°, m∠B is 60°, and m∠C is 80°. The triangle is dilated by a factor of 2 about point P to form ∆$A^{\prime} B^{\prime} C^{\prime}$. It is also dilated by a factor of 3 about point Q to form ∆$A^{\prime \prime} B^{\prime \prime} C^{\prime \prime}$ What is the measure of the angle formed by line $A^{\prime} B^{\prime}$ and line $B^{\prime \prime} C^{\prime \prime}$ ? Explain how you know.

$Eureka Math Geometry Module 2 Mid Module Assessment Answer Key 8$

Answer:

$Eureka Math Geometry Module 2 Mid Module Assessment Answer Key 17$

60˚. $\overline{B^{\prime} C^{\prime}}$ and $\overline{A^{\prime} B^{\prime}}$ meet to form a 60˚ angle. Since dilations map segments to parallel segments, $\overline{B^{\prime} C^{\prime}} \| \overline{B^{\prime \prime} C^{\prime \prime}}$. Then the angle formed by lines $A^{\prime} B^{\prime}$ and $\mathrm{B}^{\prime \prime} \mathrm{C}^{\prime \prime}$ is a corresponding angle to ∠B and has a measure of 60˚.

Question 10.
In the diagram, |AC| = |CE| = |EG|, and ∠BAC, ∠DCE, and ∠FEG are right. The two lines meet at a point to the right. Are the triangles similar? Why or why not?

$Eureka Math Geometry Module 2 Mid Module Assessment Answer Key 9$

Answer:
The triangles are not similar.
If they were similar, then there would have to be a similarity transformation taking A to C, B to D, and C to E. But then the dilation factor for this transformation would have to be 1, since AC = CE.

But since the dilation factor is 1 and $\overline{C D}$ is the image of $\overline{A B}$, then it must be true that AB = CD. Additionally, we know that $\overline{A B}$ is parallel to $\overline{C D}$, and since ∠BAC and ∠DCE are right, this implies ACDB is a rectangle. This implies that $\overline{B D}$ is parallel to $\overline{A C}$, but this is contrary to the given.

Question 11.
The side lengths of the following right triangle are 16, 30, and 34. An altitude of a right triangle from the right angle splits the hypotenuse into line segments of length x and y.

$Eureka Math Geometry Module 2 Mid Module Assessment Answer Key 10$

a. What is the relationship between the large triangle and the two sub-triangles? Why?
Answer:
An altitude drawn from the vertex of the right angle of a right triangle to the hypotenuse divides the right triangle into two sub-triangles that are similar to the original triangle by the AA criterion.

b. Solve for h, x, and y.
Answer:
$\frac{h}{30}=\frac{16}{34}$
h = $\frac{240}{17}$

$\frac{x}{16}=\frac{16}{34}$
x = $\frac{128}{17}$

$\frac{y}{30}=\frac{30}{34}$
y = $\frac{450}{17}$

c. Extension: Find an expression that gives h in terms of x and y.
Answer:
The large triangle is similar to both sub-triangles, and both sub-triangles are, therefore, similar. Then,
$\frac{x}{h}=\frac{h}{y}$
h² = xy
h = $\sqrt{x y}$

Question 12.
The sentence below, as shown, is being printed on a large banner for a birthday party. The height of the banner is 18 inches. There must be a minimum 1-inch margin on all sides of the banner. Use the dimensions in the image below to answer each question.

$Eureka Math Geometry Module 2 Mid Module Assessment Answer Key 11$

a. Describe a reasonable figure in the plane to model the printed image.
Answer:
The sentence can be modeled as a rectangle with dimensions 4.78″ × 0.44″.

b. Find the scale factor that maximizes the size of the characters within the given constraints.
Answer:
The scaled height of the rectangle cannot exceed 16″ to allow for 1″ margins above and below on the banner.
16 = k(0.44), where k represents the scale factor of the banner;
k = $\frac{16}{0.44}$ = $\frac{400}{11}$ = 36$\frac{4}{11}$

c. Wnat is the total length ot tne banner asea on your answer to part (a)?
Answer:
Using the scale factor from part (a),
y = ($\frac{400}{11}$) . (4.78)
y = 173$\frac{9}{11}$
The total length of the image in the banner is 173$\frac{9}{11}$ inches; however, the banner must
have a minimum 1″ margin on all sides of the image, so the banner must be at least 2″ longer. The total length of the banner must be at least 175$\frac{9}{11}$ inches.

Eureka Math Geometry Module 2 Lesson 20 Answer Key

May 7, 2021

Engage NY Eureka Math Geometry Module 2 Lesson 20 Answer Key

Eureka Math Geometry Module 2 Lesson 20 Example Answer Key

a. If the circumference of the earth is about 25,000 miles, what is the earth’s diameter in miles?
Answer:
$\frac{25000}{\pi}$ ≈ 8000
The earth’s diameter is approximately 8,000 miles.

b. Using part (a), what is the moon’s diameter in miles?
Answer:
$\frac{2}{7} \cdot \frac{25000}{\pi}$ ≈ 2300
The moon’s diameter is approximately 2,300 miles.

c. How far away is the moon in miles?
Answer:
$108 \cdot \frac{2}{7} \cdot \frac{25000}{\pi}$ ≈ 246 000
The moon is approximately 246,000 miles from the earth.

Eureka Math Geometry Module 2 Lesson 20 Opening Exercise Answer Key

What is a solar eclipse? What is a lunar eclipse?
Answer:
A solar eclipse occurs when the moon passes between the earth and the sun, and a lunar eclipse occurs when the earth passes between the moon and the sun.

Discussion

$Eureka Math Geometry 2 Module 2 Lesson 20 Opening Exercise Answer Key 1$
Answer:
Lead students through a conversation regarding the details of solar and lunar eclipses.
→ A total solar eclipse lasts only a few minutes because the sun and moon appear to be the same size.
→ How would appearances change if the moon were closer to the earth?
The moon would appear larger.

→ The moon would appear larger, and the eclipse would last longer. What if the moon were farther away from the earth? Would we experience a total solar eclipse?
The moon would appear smaller, and it would not be possible for a total solar eclipse to occur, because the moon would appear as a dark dot blocking only part of the sun.

→ Sketch a diagram of a solar eclipse.

Students’ knowledge on eclipses varies. This is an opportunity for students to share what they know regarding eclipses. Allow a minute of discussion, and then guide them through a basic description of the moon’s shadow and how it is conical (in 3D view), but on paper in a profile view, the shadow appears as an isosceles triangle whose base coincides with the diameter of the earth. Discuss what makes the shadows of celestial bodies similar. Describe the two parts of the moon’s shadow, the umbra and penumbra.

Note that the distances are not drawn to scale in the following image.

$Eureka Math Geometry 2 Module 2 Lesson 20 Opening Exercise Answer Key 2$
→ The umbra is the portion of the shadow where all sunlight is blocked, while the penumbra is the part of the shadow where light is only partially blocked. For the purposes of our discussion today, we will be simplifying the situation and considering only the umbra.

→ What is remarkable about the full shadow caused by the eclipse? That is, what is remarkable about the umbra and the portion of the moon that is dark? Consider the relationship between the 3D and 2D image of it.
If a cone represents the portion of the moon that is dark as well as the umbra, then the part that is entirely dark in the 2D image is an isosceles triangle.

→ We assume that shadows from the moon and the earth are all similar isosceles triangles.

This is, in fact, not the case for all planetary objects, as the shadow formed depends on how far away the light source is from the celestial body and the size of the planet. For the sake of simplification as well as approximation, this assumption is made. Then, shadows created by the moon and the earth will have the following relationship:
$Eureka Math Geometry 2 Module 2 Lesson 20 Opening Exercise Answer Key 3$

→ You can imagine simulating a solar eclipse using a marble that is one inch in diameter. If you hold it one arm length away, it will easily block (more than block) the sun from one eye. Do not try this; you will damage your eye!

→ To make the marble just barely block the sun, it must be about 9 feet (i.e., 108 inches) away from your eye. So the cone of shadow behind the marble tapers to a point, which is where your eye is, and at this point, the marble just blocks out the sun. This means that the ratio of the length of the shadow of the marble to the diameter of the marble is about 108: 1.

→ In fact, by experimenting with different-size spheres, we find that this ratio holds true regardless of the size of the sphere (or circular object) as long as the sphere is at the point where it just blocks the sun from our vantage point. In other words, whether you are using a marble, a tennis ball, or a basketball to model the eclipse, the distance the sphere must be held from the eye is 108 times the diameter of the sphere.

→ We conclude that this ratio also holds for the moon and the earth during a solar eclipse. Since the moon and sun appear to be the same size in the sky (the moon just blocks the sun in a solar eclipse), we can conclude that the distance from the earth to the moon must also be roughly 108 times the diameter of the moon because the earth is at the tip of the moon’s shadow.

→ Let us now consider what is happening in a lunar eclipse. A lunar eclipse occurs when the moon passes behind the earth. Once again, the earth, sun, and moon lie on a straight line, but this time the earth is between the moon and the sun.

→ Consequently, during a lunar eclipse, the moon is still faintly visible from the earth because of light reflected off the earth.

→ Sketch a diagram of a lunar eclipse.

$Eureka Math Geometry 2 Module 2 Lesson 20 Opening Exercise Answer Key 4$

→ Based on what we know about the ratio of distances for an object to just block the sun, we can conclude that the moon must be within 108 earth diameters; if it were not within that distance, it would not pass through the earth’s shadow, and the earth could not block the sun out completely.

→ Studying total lunar eclipses was critical to finding the distance to the moon. Other types of eclipses exist, but they involve the penumbra. The following measurement required a total eclipse, or one involving the umbra.

→ By carefully examining the shadow of the earth during a total lunar eclipse, it was determined that the width, or more specifically, the diameter of the cross section of the earth’s conical shadow at the distance of the moon, is about 2$\frac{1}{2}$. moon diameters.

→ With this measurement, a diagram like the following was constructed, where the shadow of the moon was reversed along the shadow of the earth:

$Eureka Math Geometry 2 Module 2 Lesson 20 Opening Exercise Answer Key 5$

→ What length and angle relationships can we label in this diagram?

Pose the following questions one at a time, and allow students to discuss and mark their diagrams.

→ Based on what is known about the shadows of the moon and the earth, what do we know about the measures of ∠ABF and ∠BEC?
The angle measures are equal since we are assuming the shadows can be modeled by similar isosceles triangles.

→ What is the relationship between AF and AB?
Since the distance an object must be from the eye to block out the sun is 108 times the object’s diameter, the length of AB must be 108 times that of AF.

→ What is the relationship between AF and FD?
We know that the diameter of the cross section of the earth’s conical shadow at the distance of the moon is about 2$\frac{1}{2}$ moon diameters, so the length of FD is 2.5 times the length of AF.

$Eureka Math Geometry 2 Module 2 Lesson 20 Opening Exercise Answer Key 6$

→ Reversing the moon’s shadow completes parallelogram ABCD. How can we be sure that ABCD is a parallelogram?

Allow students time to discuss why this must be true, using the relationships determined in the diagram. Share out ideas
before explaining further.

→ $\overline{A D}$ must be parallel to $\overline{B C}$ by construction. The diameter of the earth is parallel to the segment formed by the moon’s diameter and the diameter of the shadow at the distance of the moon.

→ We know that mz.ABF mz.BEC since the shadows are similar triangles. This means that $\overline{A B}$ is parallel to $\overline{D C}$ (alternate interior angles).

→ Since we know that ABCD is a parallelogram, then AD = BC, or BC = 3.5 units.

→ From our work in Lesson 19, we know the Greeks had the circumference of the earth at this time, which means they also had the earth’s diameter. Then, it was known that BC ≈ 8000 miles. With this information, we
conclude:
3.5 units ≈ 8000
1 unit ≈ 2300.

→ We already know that the distance from the moon to the earth is 108 times the diameter of the moon. If the moon has a diameter of approximately 2,300 miles, then the distance from the moon to the earth is roughly (2300 × 108) miles, or 248,000 miles.

→ With some careful observation and measurement (that of a lunar eclipse) and basic geometry, Aristarchus was able to determine a fairly accurate measure of the distance from the earth to the moon.

Eureka Math Geometry Module 2 Lesson 20 Problem Set Answer Key

Question 1.
If the sun and the moon do not have the same diameter, explain how the sun’s light can be covered by the moon during a solar eclipse.
Answer:
The farther away an object is from the viewer, the smaller that object appears. The moon is closer to the earth than the sun, and it casts a shadow where it blocks some of the light from the sun. The sun is much farther away from the earth than the moon, and because of the distance, it appears much smaller in size.

Question 2.
What would a lunar eclipse look like when viewed from the moon?
Answer:
The sun would be completely blocked out by the earth for a time because the earth casts an umbra that spans a greater distance than the diameter of the moon, meaning the moon would be passing through the earth’s shadow.

Question 3.
Suppose you live on a planet with a moon, where during a solar eclipse, the moon appears to be half the diameter of the sun.
a. Draw a diagram of how the moon would look against the sun during a solar eclipse.
Answer:
Sample response:
$Eureka Math Geometry 2 Module 2 Lesson 20 Opening Exercise Answer Key 7$

b. A 1-inch diameter marble held 100 inches away on the planet barely blocks the sun. How many moon diameters away Is the moon from the planet? Draw and label a diagram to support your answer.
Answer:
If the diameter of the moon appears to be half the diameter of the sun as viewed from the planet, then the moon will not cause a total eclipse of the sun. In the diagram, PQ is the diameter of the moon, and TU is the diameter of the sun as seen from the planet at point I.
$Eureka Math Geometry 2 Module 2 Lesson 20 Opening Exercise Answer Key 8$
The diameter of the moon is represented by distance PQ. For me to view a total eclipse, where the sun is just blocked by the moon, the moon would have to be twice as wide, so TU = 2 (P Q). ∆ TIU and ∆ YI’X are both isosceles triangles, and their vertex angles are the same, so the triangles are similar by the SAS criterion.

If the triangles are similar, then their altitudes are in the same ratio as their bases. This means = $\frac{I J}{T U}=\frac{I^{\prime} A}{Y X}$. By substituting values, $\frac{I J}{2 P Q}=\frac{100}{1}$, so Ij = 200(PQ). The moon is approximately 200 moon diameters from the planet.

c. If the diameter of the moon is approximately $\frac{3}{5}$ of the diameter of the planet and the circumference of the planet is 185,000 miles, approximately how far is the moon from the planet?
Answer:
The diameter of the planet:
d_planet = $\frac{185000}{\pi}$
d_planet = $\frac{185000}{\pi}$
The diameter of the planet is approximately 58,887 miles.

The diameter of the moon:
d_moon = $\frac{3}{5}$ d_planet
d_moon = $\frac{3}{5}\left(\frac{185000}{\pi}\right)$
d_moon = $\frac{111000}{\pi}$
The diameter of the moon is approximately 35,332 miles.

The distance of the moon from the planet:
IJ = 200(PQ)
IJ = 200$\left(\frac{111000}{\pi}\right)$
IJ = $\frac{22200000}{\pi}$
The planet’s moon is approximately 7,066,479 miles from the planet.

Eureka Math Geometry Module 2 Lesson 20 Exit Ticket Answer Key

Question 1.
On Planet A, a$\frac{1}{4}$-inch diameter ball must be held at a height of 72 inches to just block the sun. If a moon orbiting
Planet A just blocks the sun during an eclipse, approximately how many moon diameters is the moon from the planet?
Answer:
The ratio of the diameter of the ball to the specified height is $\frac{\frac{1}{4}}{72}=\frac{1}{288}$. The moon’s distance in order to just block the sun would be proportional since the shadows formed are similar triangles, so the moon would orbit approximately 288 moon diameters from Planet A.

Question 2.
Planet A has a circumference of 93,480 miles. Its moon has a diameter that is approximately $\frac{1}{8}$ that of Planet A. Find the approximate distance of the moon from Planet A.
Answer:
To find the diameter of Planet A:
$\frac{93480}{\pi}$ = d_{planet A}
The diameter of Planet A is approximately 29,756 miles.

To find the diameter of the moon:
d_moon = dp1anetA
d_moon = $\frac{1}{8}\left(\frac{93480}{\pi}\right)$
d_moon = $\frac{93480}{8 \pi}=\frac{11685}{\pi}$
The diameter of the moon is approximately 3,719 miles.

To find the distance of the moon from Planet A:
distance_moon = 288(d_moon)
distance_moon = 288$\left(\frac{11685}{\pi}\right)$
distance_moon = $\frac{3365280}{\pi}$
The distance from Planet A to its moon is approximately 1,071,202 miles.

Tag: Eureka Math Geometry Module 2

Engage NY Eureka Math Geometry Module 2 Lesson 25 Answer Key

Eureka Math Geometry Module Lesson 25 Exercise Answer Key

Eureka Math Geometry Module Lesson 25 Problem Set Answer Key

Eureka Math Geometry Module Lesson 25 Exit Ticket Answer Key

Eureka Math Geometry Module Lesson 25 Exploratory Challenge Ticket Answer Key

Engage NY Eureka Math Geometry Module 2 Lesson 32 Answer Key

Eureka Math Geometry Module 2 Lesson 32 Example Answer Key

Eureka Math Geometry Module 2 Lesson 32 Opening Exercise Answer Key

Eureka Math Geometry Module 2 Lesson 32 Exercise Answer Key

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