## Engage NY Eureka Math Precalculus Module 4 Lesson 8 Answer Key

### Eureka Math Precalculus Module 4 Lesson 8 Exercise Answer Key

Exercises

Exercise 1.

Find the value of x in the figure on the left.

Answer:

sin(33°) = \(\frac{8.4}{x}\)

x = \(\frac{8.4}{\sin \left(33^{\circ}\right)}\) ≈ 15.4

Exercise 2.

Find the value of α in the figure on the right.

Answer:

sin(α) = \(\frac{4.2}{9.1}\)

α ≈ 27°

Exercise 3.

Find all of the measurements for the triangle below.

Answer:

The measure of ∠A is 62°. The length of side \(\overline{B A}\) is 6.4. The length of side \(\overline{B C}\) is 5.8.

Exercise 4.

Find the length of side \(\overline{A C}\) in the triangle below.

Answer:

We have \(\frac{A C}{\sin \left(32^{\circ}\right)}\) = \(\frac{7.5}{\sin \left(55^{\circ}\right)}\), which means AC = \(\frac{7.5 \cdot \sin \left(32^{\circ}\right)}{\sin \left(55^{\circ}\right)}\) ≈ 4.9.

Exercise 5.

A hiker at point C is 7.5 kilometers from a hiker at point B; a third hiker is at point A. Use the angles shown in the diagram below to determine the distance between the hikers at points C and A.

Answer:

In order to apply the law of sines, first we need to find the third angle. The measure of the missing angle is 86°. Now we can use the law of sines, which gives \(\frac{A C}{\sin \left(60^{\circ}\right)}\) = \(\frac{7.5}{\sin \left(86^{\circ}\right)}\), which means AC = \(\frac{7.5 \cdot \sin \left(60^{\circ}\right)}{\sin \left(86^{\circ}\right)}\) ≈ 6.5. Thus, the two hikers at points C and A are about 6.5 km apart.

Exercise 6.

Two sides of a triangle have lengths 10.4 and 6.4. The angle opposite 6.4 is 36°. What could the angle opposite 10.4 be?

Answer:

Let β represent the angle opposite 10.4. Then, we have \(\frac{10.4}{\sin (\beta)}\) = \(\frac{6.4}{\sin \left(36^{\circ}\right)}\), which gives sin β ≈ 0.955 and β = 72.8°. We could also have β = 107.2° because an angle and its supplement have the same sine value.

Exercise 7.

Two sides of a triangle have lengths 9.6 and 11.1. The angle opposite 9.6 is 59°. What could the angle opposite 11.1 be?

Answer:

Let β represent the angle opposite 11.1. Then, we have \(\frac{11.1}{\sin (\beta)}\) = \(\frac{9.6}{\sin \left(59^{\circ}\right)}\), which gives β = 82.3°. We could also have β = 97.7°.

### Eureka Math Precalculus Module 4 Lesson 8 Problem Set Answer Key

Question 1.

Let △ABC be a triangle with the given lengths and angle measurements. Find all possible missing measurements using the law of sines.

a. a = 5, m∠A = 43°, m∠B = 80°.

Answer:

b ≈ 7.22, c ≈ 6.15, m∠C = 57°

b. a = 3.2, m∠A = 110°, m∠B = 35°.

Answer:

b ≈ 1.95, c ≈ 1.95, m∠C = 35°

c. a = 9.1, m∠A = 70°, m∠B = 95°.

Answer:

b ≈ 9.65, c ≈ 2.51, m∠C = 15°

d. a = 3.2, m∠B = 30°, m∠C = 45°.

Answer:

m∠A = 105°, b ≈ 1.66, c ≈ 2.34

e. a = 12, m∠B = 29°, m∠C = 31°.

Answer:

m∠A = 120°, b ≈ 6.72, c ≈ 7.14

f. a = 4.7, m∠B = 18.8°, m∠C = 72°.

Answer:

m∠A = 89.2°, b ≈ 1.51, c ≈ 4.47

g. a = 6, b = 3, m∠A = 91°.

Answer:

m∠B ≈ 29.99°, m∠C ≈ 59.01°, c ≈ 5.14

h. a = 7.1, b = 7, m∠A = 70°.

Answer:

m∠B = 67.89°, m∠C ≈ 42.11°, c = 5.07

i. a = 8, b = 5, m∠A = 45°.

Answer:

m∠B = 26.23°, m∠C = 108.77°, c = 10.71

j. a = 3.5, b = 3.6, m∠A = 37°.

Answer:

m∠B = 38.24°, m∠C = 104.76°, c = 5.62 or m∠B = 141.76°, m∠C = 1.24°, c = 0.13

k. a = 9, b = 10.1, m∠A = 61°.

Answer:

m∠B = 78.97°, m∠C = 40.03°, c = 6.62 or m∠B = 101.03°, m∠C = 17.97°, c = 3.17

l. a = 6, b = 8, m∠A = 41.5°.

Answer:

m∠B = 62.07°, m∠C = 76.43°, c = 8.8 or m∠B = 117.93°, m∠C = 20.57°, c = 3.18

Question 2.

A surveyor is working at a river that flows north to south. From her starting point, she sees a location across the river that is 20° north of east from her current position, she labels the position S. She moves 110 feet north and measures the angle to S from her new position, seeing that it is 32° south of east.

a. Draw a picture representing this situation.

Answer:

b. Find the distance from her starting position to S.

Answer:

\(\frac{\sin \left(52^{\circ}\right)}{110}\) = \(\frac{\sin \left(58^{\circ}\right)}{x}\)

x ≈ 118.4

The distance is about 118 ft.

c. Explain how you can use the procedure the surveyor used in this problem (called triangulation) to calculate the distance to another object.

Answer:

Calculate the angle from the starting point to the object. Travel a distance from the starting point, and again calculate the angle to the object. Create a triangle connecting the starting point, the object, and the new location. Use that distance traveled and the angles to find the distance to the object.

Question 3.

Consider the triangle pictured below.

Use the law of sines to prove the generalized angle bisector theorem, that is, \(\frac{BD}{DC}\) = \(\frac{1}{2}\). (Although this is called the generalized angle bisector theorem, we do not assume that the bisector of ∠BAC intersects side

\(\overline{B C}\) at D. In the case that \(\overline{A D}\) is an angle bisector, then the formula simplifies to \(\frac{BD}{DC}\) = \(\frac{c}{b}\).)

a. Use the triangles ABD and ACD to express \(\frac{c}{BD}\) and \(\frac{b}{DC}\) as a ratio of sines.

Answer:

\(\frac{c}{BD}\) = \(\frac{\sin (m \angle B D A)}{\sin (m \angle B A D)}\)

\(\frac{b}{DC}\) = \(\frac{\sin (m \angle C D A)}{\sin (m \angle C A D)}\)

b. Note that angles BDA and ADC form a linear pair. What does this tell you about the value of the sines of these angles?

Answer:

Since the angles are supplementary, the sines of these values are equal.

c. Solve each equation in part (a) to be equal to the sine of either ∠BDA or ∠ADC.

Answer:

sin(m∠BAD)⋅\(\frac{c}{BD}\) = sin(m∠BDA)

sin(m∠CAD)⋅\(\frac{b}{DC}\) = sin(m∠CDA)

d. What do your answers to parts (b) and (c) tell you?

Answer:

The answers tell me that the two equations written in part (c) are equal to each other.

e. Prove the generalized angle bisector theorem.

Answer:

From part (d), we have

sin(m∠BAD)⋅\(\frac{c}{BD}\) = sin(m∠CAD)⋅\(\frac{b}{DC}\)

Dividing both sides by sin(m∠CAD)⋅b, and multiplying by BD, we get

\(\frac{BD}{DC}\) = \(\frac{c \sin (m \angle B A D)}{b \sin (m \angle C A D)}\).

Question 4.

As an experiment, Carrie wants to independently confirm the distance to Alpha Centauri. She knows that if she measures the angle of Alpha Centauri and waits 6 months and measures again, then she will have formed a massive triangle with two angles and the side between them being 2 AU long.

a. Carrie measures the first angle at 82° 8′ 24.5” and the second at 97° 51′ 34”. How far away is Alpha Centauri according to Carrie’s measurements?

Answer:

The third angle would be 1.5′.

\(\frac{\sin \left(82+\frac{8}{60}+\frac{24.5}{3600}\right)}{a}\) = \(\frac{\sin \left(\frac{1.5}{3600}\right)}{2}\)

a ≈ 272 436

About 272,436 AU.

b. Today, astronomers use the same triangulation method on a much larger scale by finding the distance between different spacecraft using radio signals and then measuring the angles to stars. Voyager 1 is about 122 AU away from Earth. What fraction of the distance from Earth to Alpha Centauri is this? Do you think that measurements found in this manner are very precise?

Answer:

Voyager 1 is about \(\frac{122}{276364}\), or 0.0004 the distance of Earth to Alpha Centauri. Depending on how far away the object being measured is, the distances are fairly precise on an astronomical scale. One AU is almost

93 million miles, which is not very precise.

Question 5.

A triangular room has sides of length 3.8 m, 5.1 m, and 5.1 m. What is the area of the room?

Answer:

Since the room is isosceles, the height bisects the side of length 3.8 at a right angle. We get cos(θ) = \(\frac{1.9}{5.1}\); therefore, θ ≈ 68.127°.

\(\frac{1}{2}\)⋅3.8⋅5.1⋅sin(68.127°) ≈ 8.99

The area of the room is approximately 8.99 m^{2}.

Question 6.

Sara and Paul are on opposite sides of a building that a telephone pole fell on. The pole is leaning away from Paul at an angle of 59° and towards Sara. Sara measures the angle of elevation to the top of the telephone pole to be 22°, and Paul measures the angle of elevation to be 34°. Knowing that the telephone pole is about 35 ft. tall, answer the following questions.

a. Draw a diagram of the situation.

Answer:

b. How far apart are Sara and Paul?

Answer:

We can use law of sines to find the shared side between the two triangles and then again with the larger triangle to find the distance:

\(\frac{\sin \left(22^{\circ}\right)}{35}\) = \(\frac{\sin \left(59^{\circ}\right)}{r}\)

r ≈ 80.086

Then we use this value in the larger triangle:

\(\frac{\sin \left(124^{\circ}\right)}{x}\) = \(\frac{\sin \left(34^{\circ}\right)}{80.086}\)

x ≈ 118.733

They are about 118.7 ft. away from each other.

c. If we assume the building is still standing, how tall is the building?

Answer:

sin(59°) = \(\frac{y}{35}\)

y ≈ 30.001

The building is about 30 ft. tall.

### Eureka Math Precalculus Module 4 Lesson 8 Exit Ticket Answer Key

Question 1.

Find the length of side \(\overline{A C}\) in the triangle below.

Answer:

We have \(\frac{A C}{\sin \left(105^{\circ}\right)}\) = \(\frac{11.7}{\sin \left(41^{\circ}\right)}\), which gives C = \(\frac{11.7 \cdot \sin \left(105^{\circ}\right)}{\sin \left(41^{\circ}\right)}\) ≈ 17.2 .

Question 2.

A triangle has sides with lengths 12.6 and 7.9. The angle opposite 7.9 is 37°. What are the possible values of the measure of the angle opposite 12.6?

Answer:

We have \(\frac{\sin (x)}{12.6}\) = \(\frac{\sin \left(37^{\circ}\right)}{7.9}\), which means that sin(x) = \(\frac{12.6 \cdot \sin \left(37^{\circ}\right)}{7.9}\). The values of x that satisfy this equation are x ≈ 73.7° and x ≈ 106.3°.