Engage NY Eureka Math Geometry Module 2 Lesson 29 Answer Key
Eureka Math Geometry Module 2 Lesson 29 Example Answer Key
Example 1.
Scott, whose eye level is 1.5 m above the ground, stands 30 m from a tree. The angle of elevation of a bird at the top of the tree is 36°. How far above ground is the bird?
Answer:
→ With respect to your diagram, think of the measurement you are looking for.
In our diagram, we are looking for BC.
→ How will you find BC?
I can use the tangent to determine BC in meters.
tan 36 = \(\frac{B C}{A C}\)
tan 36 = \(\frac{B C}{30}\)
30 tan 36 = BC
BC ≈ 21.8
→ Have we found the height at which the bird is off the ground?
No. The full height must be measured from the ground, so the distance from the ground to Scott’s eye level must be added to BC.
The height of the bird off of the ground is 1.5 m + 21.8 m = 23.3 m.
→ So, between the provided measurements, including the angle of elevation, and the use of the tangent ratio, we were able to find the height of the bird.
Example 2
From an angle of depression of 40°, John watches his friend approach his building while standing on the rooftop. The rooftop is 16 m from the ground, and John’s eye level is at about 1.8 m from the rooftop. What is the distance between John’s friend and the building?
Answer:
Make sure to point out the angle of depression in the diagram below. Emphasize that the 40° angle of depression is the angle between the line of sight and the line horizontal (to the ground) from the eye.
→ Use the diagram to describe the distance we must determine.
We are going to find BC in meters.
→ How will we find BC?
We can use the tangent: tan 40 = \(\frac{A B}{B C}\)
tan 40 = \(\frac{A B}{B C}\)
tan 40 = \(\frac{17.8}{B C}\)
BC = \(\)
BC ≈ 21.2
→ Again, with the assistance of a few measurements, including the angle of depression, we were able to determine the distance between John’s friend and the building, which is 21.2 meters.
Eureka Math Geometry Module 2 Lesson 29 Opening Exercise Answer Key
a. Use a calculator to find the tangent of θ. Enter the values, correct to four decimal places, in the last row of the table.
Answer:
Note to the teacher: Dividing the values in the first two rows provides a different answer than using the full decimal readout of the calculator.
b. The table from Lesson 29 is provided here for you. In the row labeled \(\frac{\sin \theta}{\cos \theta}\), divide the sine values by the cosine values. What do you notice?
Answer:
For each of the listed degree values, dividing the value of sin θ by the corresponding value of cos θ yields a value very close to the value of tan θ. The values divided to obtain \(\frac{\sin \theta}{\cos \theta}\) were approximations, so the actual
values might be exactly the same.
Eureka Math Geometry Module 2 Lesson 29 Exercise Answer Key
Exercise 1.
Standing on the gallery of a lighthouse (the deck at the top of a lighthouse), a person spots a ship at an angle of depression of 20°. The lighthouse is 28 m tall and sits on a cliff 45 m tall as measured from sea level. What is the horizontal distance between the lighthouse and the ship? Sketch a diagram to support your answer.
Answer:
Approximately 201 m
Exercise 2.
A line on the coordinate plane makes an angle of depression of 36°. Find the slope of the line correct to four decimal places.
Answer:
Choose a segment on the line, and construct legs of a right triangle parallel to the x- and y-axes, as shown. If m is the length of the vertical leg and n is the length of the horizontal leg, then tan 36 = \(\frac{m}{n}\). The line decreases to the right, so the value of the slope must be negative. Therefore,
slope = –\(\frac{m}{n}\) = – tan 36 ≈ – 0.7265.
Eureka Math Geometry Module 2 Lesson 29 Problem Set Answer Key
Question 1.
A line in the coordinate plane has an angle of elevation of 53°. Find the slope of the line correct to four decimal places.
Answer:
Since parallel lines have the same slope, we can consider the line that passes through the origin with an angle of inclination of 53°. Draw a vertex at (5, 0) on the x-axis, and draw a segment from (5, 0) parallel to the y-axis to the intersection with the line to form a right triangle. If the base of the right triangle is 5-units long, let the height of the triangle be represented by y.
tan 53 = \(\frac{x}{5}\)
5(tan 53) = x
The slope of the line is
slope = tan 53 ≈ 1.3270.
Question 2.
A line in the coordinate plane has an angle of depression of 25°. Find the slope of the line correct to four decimal places.
Answer:
A line crosses the x-axis, decreasing to the right with an angle of depression of 25°. Using a similar method as in Problem 1, a right triangle is formed. If the leg along the x-axis represents the base of the triangle, let h represent the height of the triangle.
Using tangent:
tan 25 = \(\frac{h}{5}\)
5(tan 25) = h
The slope of the line is negative since the line decreases to the right:
Question 3.
In Problems 1 and 2, why do the lengths of the legs of the right triangles formed not affect the slope of the line?
Answer:
When using the tangent ratio, the length of one leg of a right triangle can be determined in terms of the other leg. Let x represent the length of the horizontal leg of a slope triangle. The vertical leg of the triangle is then x tan θ, where θ is the measure of the angle of inclination or depression. The slope of the line is
Question 4.
Given the angles of depression below, determine the slope of the line with the Indicated angle correct to four decimal places.
a. 35° angle of depression
Answer:
tan 35 ≈ 0.7002
slope ≈ – 0.7002
b. 49° angle of depression
Answer:
tan 49 ≈ 1. 1504
slope ≈ – 1.1504
c. 80° angle of depression
Answer:
tan 80 ≈ 5. 6713
slope ≈ – 5. 6713
d. 87° angle of depression
Answer:
tan 87 ≈ 19. 0811
slope ≈ – 19. 0811
e. 89° angle of depression
Answer:
tan 89 ≈ 57. 2900
slope ≈ – 57.2900
f. 89.9° angle of depression
Answer:
tan 89.9 ≈ 572.9572
slope ≈ – 572.9572
g. What appears to be happening to the slopes (and tangent values) as the angles of depression get closer to 90°?
Answer:
As the angles get closer to 90°, their slopes (and tangent values) get much farther from zero.
h. Find the slopes of angles of depression that are even closer to 90° than 89.90. Can the value of the tangent of 90° be defined? Why or why not?
Answer:
Choices of angles will vary. The closer an angle is in measure to 90°, the greater the tangent value of that angle and the farther the slope of the line determined by that angle is from zero. An angle of depression of 90° would be a vertical line, and vertical lines have 0 run; therefore, the value of the ratio rise: run is undefined. The value of the tangent of 90° would have a similar outcome because the adjacent leg of the “triangle” would have a length of 0, so the ratio \(\frac{\text { opp }}{\text { adj }}=\frac{\text { opp }}{0}\), which is undefined.
Question 5.
For the indicated angle, express the quotient in terms of sine, cosine, or tangent. Then, write the quotient in simplest terms.
a. \(\frac{4}{2 \sqrt{13}}\); α
Answer:
cos α = \(\frac{4}{2 \sqrt{13}}=\frac{2}{\sqrt{13}}=\frac{2 \sqrt{13}}{13}\)
b. \(\frac{6}{4}\); α
Answer:
tan α = \(\frac{6}{4}=\frac{3}{2}\)
c. \(\frac{4}{2 \sqrt{13}}\); β
Answer:
sin β = \(\frac{4}{2 \sqrt{13}}=\frac{2 \sqrt{13}}{13}\)
d. \(\frac{4}{6}\); β
Answer:
tan β = \(\frac{4}{6}=\frac{2}{3}\)
Question 6.
The pitch of a roof on a home Is expressed as a ratio of vertical rise: horizontal run where the run has a length of 12 units. If a given roof design includes an angle of elevation of 22. 5° and the roof spans 36 ft. as shown in the diagram, determine the pitch of the roof. Then, determine the distance along one of the two sloped surfaces of the roof.
Answer:
The diagram as shown is an isosceles triangle since the base angles have equal measure. The altitude, a, of the triangle is the vertical rise of the roof.
The right triangles formed by drawing the altitude of the given isosceles triangle have a leg of length 18 ft.
tan 225 = \(\frac{a}{18}\)
a = 18 tan 22.5
a ≈ 7.5
Roof pitch:
\(\frac{7.5}{18}=\frac{h}{12}\)
18h = 12 · 7.5
h = 5
cos 22.5 = \(\frac{18}{s}\)
s = \(\frac{18}{\cos 22.5}\)
s ≈ 19.5
The pitch of the roof is 5:12.
The sloped surface of the roof has a distance of approximately 19.5 ft.
Question 7.
An anchor cable supports a vertical utility pole forming a 51° angle with the ground. The cable is attached to the top of the pole. If the distance from the base of the pole to the base of the cable is 5 meters, how tall is the pole?
Answer:
Let h represent the height of the pole in meters.
Using tangent:
tan 51 = \(\frac{h}{5}\)
5(tan 51) = h
6.17 ≈ h
The height of the utility pole is approximately 6. 17 meters.
Question 8.
A winch is a tool that rotates a cylinder, around which a cable is wound. When the winch rotates in one direction, it draws the cable in. Joey is using a winch and a pulley (as shown in the diagram) to raise a heavy box off the floor and onto a cart. The box is 2 ft. tall, and the winch is 14 ft. horizontally from where cable drops down vertically from the pulley. The angle of elevation to the pulley is 42°. What is the approximate length of cable required to connect the winch and the box?
Answer:
Let h represent the length of cable in the distance from the winch to the pulley along the hypotenuse of the right triangle shown in feet, and let y represent the distance from the pulley to the floor in feet.
Using tangent:
tan 42 = \(\frac{y}{14}\)
14(tan 42) = y
12.61 ≈ y
Using cosine:
cos 42 = \(\frac{14}{h}\)
h = \(\frac{14}{\cos 42}\)
h ≈ 18.84
Let t represent the total amount of cable from the winch to the box in feet:
t ≈ 12.61 + 18.84 – 2
t ≈ 29.45
The total length of cable from the winch to the box is approximately 29.45 ft.
Eureka Math Geometry Module 2 Lesson 29 Exit Ticket Answer Key
Question 1.
A line on the coordinate plane makes an angle of depression of 24°. Find the slope of the line correct to four decimal places.
Answer:
Choose a segment on the line, and construct legs of a right triangle parallel to the x- and y-axes, as shown. If m is the length of the vertical leg and n is the length of the horizontal leg, then tan 24 = \(\frac{m}{n}\) The line decreases to the right, so the value of the slope must be negative. Therefore,
slope = – \(\frac{m}{n}\) = – tan 24 ≈ – 0.4452.
Question 2.
Samuel is at the top of a tower and will ride a trolley down a zip line to a lower tower. The total vertical drop of the zip line is 40 ft. The zip line’s angle of elevation from the lower tower is 11. 5°. What is the horizontal distance between the towers?
Answer:
A right triangle is formed by the zip line’s path, the vertical drop along the upper tower, and the horizontal distance between the towers. Let x represent the horizontal distance between the towers in feet. Using tangent:
tan 11.5 = \(\frac{40}{x}\)
x = \(\frac{40}{\tan 11.5}\)
x ≈ 196.6
The horizontal distance between the towers ¡s approximately 196.6 ft.