A 48-foot building casts a shadow 72 feet measured along the ground. What is the angle of the sun's position above the horizon? While flying a kite, Jimmy lets out 150 feet of string. If the string makes an angle of 38 degrees with the ground, (a) how high is the kite? (b) how far from Jimmy is his friend Jane, who is directly below the kite? (Ignore the sagging of the kite string; assume the ground is level.) You take a stroll one fine evening. First you go 4.0 kilometers to the east, followed by 8.0 kilometers northward. Tired from your 12.0-km walk, you call your roommate and ask him/her to fly your shared private helicopter from your home to your location, so you can skip the walk back. How far, and in which direction, does your friend need to go to find you? Give your direction as the angle measured clockwise from due north (called a bearing).
A 48-foot building casts a shadow 72 feet measured along the ground. What is the angle of the sun's position above the horizon? While flying a kite, Jimmy lets out 150 feet of string. If the string makes an angle of 38 degrees with the ground, (a) how high is the kite? (b) how far from Jimmy is his friend Jane, who is directly below the kite? (Ignore the sagging of the kite string; assume the ground is level.) You take a stroll one fine evening. First you go 4.0 kilometers to the east, followed by 8.0 kilometers northward. Tired from your 12.0-km walk, you call your roommate and ask him/her to fly your shared private helicopter from your home to your location, so you can skip the walk back. How far, and in which direction, does your friend need to go to find you? Give your direction as the angle measured clockwise from due north (called a bearing).
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE:
1. Give the measures of the complement and the supplement of an angle measuring 35°.
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Hello, I need help solving these three questions please. Thank you.
![### Geometry and Trigonometry Example Problems
#### 3.1
A 48-foot building casts a shadow 72 feet measured along the ground. What is the angle of the sun’s position above the horizon?
To solve this problem, you can use trigonometry, specifically the tangent function, which relates angles to side lengths in a right-angled triangle.
Step-by-step solution:
1. Let θ be the angle of elevation of the sun.
2. The tangent of the angle θ (tan θ) is the ratio of the opposite side (the height of the building) to the adjacent side (the length of the shadow).
\[
\tan(θ) = \frac{{48}}{{72}}
\]
3. Calculate the ratio:
\[
\tan(θ) = \frac{48}{72} = \frac{2}{3}
\]
4. Use the inverse tangent function (arctan or tan^(-1)) to find the angle θ:
\[
θ = \tan^{-1}\left(\frac{2}{3}\right)
\]
5. θ ≈ 33.69 degrees
Therefore, the angle of the sun’s position above the horizon is approximately 33.69 degrees.
#### 3.2
While flying a kite, Jimmy lets out 150 feet of string. If the string makes an angle of 38 degrees with the ground, (a) how high is the kite? (b) how far from Jimmy is his friend Jane, who is directly below the kite? (Ignore the sagging of the kite string; assume the ground is level.)
Step-by-step solutions:
**(a)** How high is the kite?
1. Let h be the height of the kite.
2. Use the sine function, which relates the opposite side (height of the kite) to the hypotenuse (length of the string):
\[
\sin(38^{\circ}) = \frac{h}{150}
\]
3. Rearrange to solve for h:
\[
h = 150 \times \sin(38^{\circ})
\]
4. Calculate h:
\[
h ≈ 150 \times 0.6157 ≈ 92.36 \text{ feet}
\]
**(b)** How far from Jimmy is his friend Jane, who is directly](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb3524539-73d7-4b1e-9607-006acd5ee568%2F0b0e6527-5e16-461d-a08d-1865c013457f%2Fbru3gb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Geometry and Trigonometry Example Problems
#### 3.1
A 48-foot building casts a shadow 72 feet measured along the ground. What is the angle of the sun’s position above the horizon?
To solve this problem, you can use trigonometry, specifically the tangent function, which relates angles to side lengths in a right-angled triangle.
Step-by-step solution:
1. Let θ be the angle of elevation of the sun.
2. The tangent of the angle θ (tan θ) is the ratio of the opposite side (the height of the building) to the adjacent side (the length of the shadow).
\[
\tan(θ) = \frac{{48}}{{72}}
\]
3. Calculate the ratio:
\[
\tan(θ) = \frac{48}{72} = \frac{2}{3}
\]
4. Use the inverse tangent function (arctan or tan^(-1)) to find the angle θ:
\[
θ = \tan^{-1}\left(\frac{2}{3}\right)
\]
5. θ ≈ 33.69 degrees
Therefore, the angle of the sun’s position above the horizon is approximately 33.69 degrees.
#### 3.2
While flying a kite, Jimmy lets out 150 feet of string. If the string makes an angle of 38 degrees with the ground, (a) how high is the kite? (b) how far from Jimmy is his friend Jane, who is directly below the kite? (Ignore the sagging of the kite string; assume the ground is level.)
Step-by-step solutions:
**(a)** How high is the kite?
1. Let h be the height of the kite.
2. Use the sine function, which relates the opposite side (height of the kite) to the hypotenuse (length of the string):
\[
\sin(38^{\circ}) = \frac{h}{150}
\]
3. Rearrange to solve for h:
\[
h = 150 \times \sin(38^{\circ})
\]
4. Calculate h:
\[
h ≈ 150 \times 0.6157 ≈ 92.36 \text{ feet}
\]
**(b)** How far from Jimmy is his friend Jane, who is directly
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