## 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 âˆ BAC and âˆ ACB are 41Â° and 55Â°, respectively. Approximate the distance from A to B to the nearest meter.

Answer:

â†’ What measurement can we add to the diagram based on the provided information?

The measurement of âˆ B must be 84Â° by the triangle sum theorem.

â†’ Use the law of sines to set up all possible ratios applicable to the diagram.

\(\frac{\sin 41}{a}\) = \(\frac{\sin 84}{160}\) = \(\frac{\sin 55}{c}\)

â†’ Which ratios will be relevant to determining the distance from A to B?

\(\frac{\sin 84}{160}\) = \(\frac{\sin 55}{c}\)

â†’ 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 âˆ ABD and âˆ BAD both measure 75Â°. What is the distance between D and A to the nearest meter?

Answer:

â†’ What do you notice about âˆ† ABD right away?

âˆ† ABD must be an isosceles triangle since it has two angles of equal measure.

â†’ 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.

â†’ 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.

b^{2} = 132^{2} + a^{2} – 2(132)(a) cos 75

â†’ 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:

b^{2} = 132^{2} + (b)^{2} – 2(132) (b) cos 75

b^{2} = 132^{2} + (b)^{2} – 264(b) cos 75

0 = 132^{2} – 264(b)cos 75

264(b) cos 75 = 132^{2}

b = \(\frac{132^{2}}{264 \cos 75}\)

b â‰ˆ 255 m

### Eureka Math Geometry Module 2 Lesson 32 Opening Exercise Answer Key

a. Find the length of d and e.

Answer:

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

Answer:

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

### Eureka Math Geometry Module 2 Lesson 32 Exercise Answer Key

Exercise 1.

In âˆ† ABC, mâˆ A 30, a = 12, and b = 10. Find sinâˆ B. Include a diagram in your answer.

Answer:

\(\frac{\sin 30}{12}=\frac{\sin \angle B}{10}\)

sin âˆ B = \(\frac{5}{12}\)

Exercise 2.

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

Answer:

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

\(\overline{B H}\) is the hypotenuse of a 45 – 45 – 90 triangle whose sides are in the ratio 1: 1: âˆš2, or h: h: hâˆš2.

hâˆš2 = x

hâˆš2 = \(\frac{125}{\sin 15}\)

h = \(\frac{125}{\sin 15 \cdot \sqrt{2}}\)

h â‰ˆ 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 100Â°.

Approximate the length of diagonal \(\overline{A C}\) to the nearest millimeter.

Answer:

In parallelogram ABCD, mâˆ C = 100Â°; therefore, mâˆ D = 80Â°.

Let d represent the length of \(\overline{A C}\).

d^{2} = 44^{2} + 26^{2} – 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 âˆ† ABC, AB = 14, mâˆ A = 57. 2Â°, and mâˆ C = 78. 4Â°, calculate the measure of angle B to the nearest tenth of a degree, and use the law of sines to find the lengths of \(\overline{A C}\) and \(\overline{B C}\) to the nearest tenth.

Answer:

By the angle sum of a triangle, mâˆ B = 44.4Â°.

Calculate the area of âˆ† 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 â‰ˆ 59

Question 2.

Given âˆ† DEF, mâˆ F = 39Â°, and EF = 13, calculate the measure of âˆ E, and use the law of sines to find the lengths of \(\overline{D F}\) and \(\overline{D E}\) to the nearest hundredth.

Answer:

By the angle sum of a triangle, mâˆ E = 55Â°.

Question 3.

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

Fill in the partially completed work below:

What conclusions can we draw?

Answer:

The law of sines does apply to a right triangle. We get the formulas that are equivalent to sin âˆ A = \(\frac{\text { opp }}{\text { hyp }}\) and sin âˆ 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, mâˆ H = 50Â°, mâˆ HKG = 80Â°, mLKGJ = 50Â°, âˆ j is a right angle, and GH = 9 in, use the law of sines to find the length of \(\overline{G K}\), and then find the lengths of \(\overline{G J}\) and \(\overline{J K}\) to the nearest tenth of an inch.

Answer:

By the angle sum of a triangle, mâˆ HGK = 50Â°; therefore,

âˆ† GHK is an isosceles triangle since its base âˆ â€˜s have equal measure.

\(\frac{\sin 50}{h}=\frac{\sin 80}{9}\)

h = \(\frac{9 \sin 50}{\sin 80}\)

h â‰ˆ 7.0

k = 7 cos 50 â‰ˆ 4.5

g = 7 sin 50 â‰ˆ 5.4

Question 5.

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

Answer:

l^{2} = 10^{2} + 15^{2} – 2(10)(15) cos 38

l^{2} = 100 + 225 – 300 cos 38

l^{2} = 325 – 300 cos 38

l^{2} = 325 – 300 cos 38

l = \(\sqrt{325-300 \cos 38}\)

l â‰ˆ 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 mâˆ A = 78Â°, draw a diagram of triangle ABC, and use the law of cosines to find the length \(\overline{B C}\).

Answer:

a^{2} = 6^{2} + 8^{2} – 2(6)(8)(cos 78)

a^{2} = 36 + 64 – 96(cos 78)

a^{2} = 100 – 96 cos 78

a = \(\sqrt{100-96 \cos 78}\)

a â‰ˆ 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 â‰ˆ 23.5

The area of triangle ABC is approximately 23.5 square units.

### Eureka Math Geometry Module 2 Lesson 32 Exit Ticket Answer Key

Question 1.

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

Answer:

Question 2.

Given âˆ† DEF, use the law of cosines to find the length of the side marked d to the nearest tenth.

Answer:

d^{2} = 6^{2} + 9^{2} – 2(6)(9) (cos 65)

d^{2} = 36 + 81 – 108(cos 65)

d^{2} = 117 – 108 (cos 65)

d = \(\sqrt{117-108(\cos 65)}\)

d â‰ˆ 8.4