Solved Name Date Angle of Elevation & Depression Worksheet | Chegg.com - Free Printable

Problem 5:

A man flies a kite with a 100-foot string. The angle of elevation of the string is \(52^\circ\). How high off the ground is the kite?

#### Solution:
We are given:
- Length of the string (hypotenuse) = 100 feet
- Angle of elevation = \(52^\circ\)
- We need to find the height of the kite above the ground, which corresponds to the opposite side of the right triangle.

Using trigonometry, specifically the sine function:
\[
\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}
\]
Here, \(\theta = 52^\circ\), the opposite side is the height of the kite (\(x\)), and the hypotenuse is 100 feet. So:
\[
\sin(52^\circ) = \frac{x}{100}
\]
Solving for \(x\):
\[
x = 100 \cdot \sin(52^\circ)
\]
Using a calculator to find \(\sin(52^\circ)\):
\[
\sin(52^\circ) \approx 0.788
\]
Thus:
\[
x = 100 \cdot 0.788 = 78.8
\]
Rounding to the nearest tenth:
\[
x \approx 78.8 \text{ feet}
\]

#### Final Answer:
\[
\boxed{78.8}
\]

---

Problem 6:

From the top of a vertical cliff 40 meters high, the angle of depression of an object that is level with the base of the cliff is \(34^\circ\). How far is the object from the base of the cliff?

#### Solution:
We are given:
- Height of the cliff = 40 meters
- Angle of depression = \(34^\circ\)
- We need to find the horizontal distance from the base of the cliff to the object.

The angle of depression is the same as the angle of elevation from the object to the top of the cliff. So, we can use the tangent function:
\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
\]
Here, \(\theta = 34^\circ\), the opposite side is the height of the cliff (40 meters), and the adjacent side is the horizontal distance (\(x\)). So:
\[
\tan(34^\circ) = \frac{40}{x}
\]
Solving for \(x\):
\[
x = \frac{40}{\tan(34^\circ)}
\]
Using a calculator to find \(\tan(34^\circ)\):
\[
\tan(34^\circ) \approx 0.675
\]
Thus:
\[
x = \frac{40}{0.675} \approx 59.26
\]
Rounding to the nearest tenth:
\[
x \approx 59.3 \text{ meters}
\]

#### Final Answer:
\[
\boxed{59.3}
\]

---

Problem 7:

An airplane takes off 200 yards in front of a 60-foot building. At what angle of elevation must the plane take off in order to avoid crashing into the building? Assume that the airplane flies in a straight line and the angle of elevation remains constant until the airplane flies over the building.

#### Solution:
We are given:
- Distance from the airplane to the building = 200 yards = \(200 \times 3 = 600\) feet (since 1 yard = 3 feet)
- Height of the building = 60 feet
- We need to find the angle of elevation (\(\theta\)).

Using the tangent function:
\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
\]
Here, the opposite side is the height of the building (60 feet), and the adjacent side is the horizontal distance (600 feet). So:
\[
\tan(\theta) = \frac{60}{600} = 0.1
\]
To find \(\theta\), we take the inverse tangent:
\[
\theta = \tan^{-1}(0.1)
\]
Using a calculator:
\[
\theta \approx 5.71^\circ
\]
Rounding to the nearest tenth:
\[
\theta \approx 5.7^\circ
\]

#### Final Answer:
\[
\boxed{5.7}
\]

---

Problem 8:

A 14-foot ladder is used to scale a 13-foot wall. At what angle of elevation must the ladder be situated in order to reach the top of the wall?

#### Solution:
We are given:
- Length of the ladder (hypotenuse) = 14 feet
- Height of the wall (opposite side) = 13 feet
- We need to find the angle of elevation (\(\theta\)).

Using the sine function:
\[
\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}
\]
Here, the opposite side is the height of the wall (13 feet), and the hypotenuse is the length of the ladder (14 feet). So:
\[
\sin(\theta) = \frac{13}{14}
\]
To find \(\theta\), we take the inverse sine:
\[
\theta = \sin^{-1}\left(\frac{13}{14}\right)
\]
Using a calculator:
\[
\theta \approx \sin^{-1}(0.9286) \approx 68.2^\circ
\]
Rounding to the nearest tenth:
\[
\theta \approx 68.2^\circ
\]

#### Final Answer:
\[
\boxed{68.2}
\]

---

Summary of Answers:

1. Problem 5: \(\boxed{78.8}\)
2. Problem 6: \(\boxed{59.3}\)
3. Problem 7: \(\boxed{5.7}\)
4. Problem 8: \(\boxed{68.2}\)