Worksheet on angles of depression and elevation with four illustrated trigonometry problems.

Problem 1:

A kite with a string 150 feet long makes an angle of 45° with the ground. Assuming the string is straight, how high is the kite?

#### Solution:
We are given:
- Length of the string (hypotenuse) = 150 feet
- Angle between the string and the ground = 45°

We need to find the height of the kite, which corresponds to the opposite side of the right triangle formed by the string, the ground, and the vertical height of the kite.

Using trigonometry, specifically the sine function:
\[
\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}
\]
Here, \(\theta = 45^\circ\), the opposite side is the height of the kite (\(x\)), and the hypotenuse is 150 feet. So:
\[
\sin(45^\circ) = \frac{x}{150}
\]

We know that \(\sin(45^\circ) = \frac{\sqrt{2}}{2}\). Substituting this value:
\[
\frac{\sqrt{2}}{2} = \frac{x}{150}
\]

Solving for \(x\):
\[
x = 150 \cdot \frac{\sqrt{2}}{2} = 150 \cdot 0.7071 \approx 106.07 \text{ feet}
\]

Thus, the height of the kite is:
\[
\boxed{106.07 \text{ feet}}
\]

---

Problem 2:

A tree 10 meters high casts a 17.3-meter shadow. Find the angle of elevation of the sun.

#### Solution:
We are given:
- Height of the tree (opposite side) = 10 meters
- Length of the shadow (adjacent side) = 17.3 meters

We need to find the angle of elevation of the sun, which is the angle \(\theta\) between the ground and the line from the top of the tree to the tip of the shadow.

Using trigonometry, specifically the tangent function:
\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
\]
Here, the opposite side is the height of the tree (10 meters), and the adjacent side is the length of the shadow (17.3 meters). So:
\[
\tan(\theta) = \frac{10}{17.3}
\]

Calculating the value:
\[
\tan(\theta) \approx 0.578
\]

To find \(\theta\), we take the inverse tangent (arctangent):
\[
\theta = \arctan(0.578)
\]

Using a calculator:
\[
\theta \approx 30^\circ
\]

Thus, the angle of elevation of the sun is:
\[
\boxed{30^\circ}
\]

---

Problem 3:

A plane is flying at an altitude of 12,000 meters. From the pilot, the angle of depression to the airport tower is 32°. How far is the tower from a point directly beneath the plane?

#### Solution:
We are given:
- Altitude of the plane (opposite side) = 12,000 meters
- Angle of depression = 32°

The angle of depression is the same as the angle of elevation from the ground to the plane. We need to find the horizontal distance (\(x\)) from the point directly beneath the plane to the airport tower.

Using trigonometry, specifically the tangent function:
\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
\]
Here, \(\theta = 32^\circ\), the opposite side is the altitude of the plane (12,000 meters), and the adjacent side is the horizontal distance (\(x\)). So:
\[
\tan(32^\circ) = \frac{12,000}{x}
\]

Solving for \(x\):
\[
x = \frac{12,000}{\tan(32^\circ)}
\]

Using a calculator to find \(\tan(32^\circ)\):
\[
\tan(32^\circ) \approx 0.6249
\]

Substituting this value:
\[
x = \frac{12,000}{0.6249} \approx 19,208.6 \text{ meters}
\]

Thus, the distance from the point directly beneath the plane to the airport tower is:
\[
\boxed{19,209 \text{ meters}}
\]

---

Problem 4:

A car is traveling up a slight grade with an angle of elevation of 2°. After traveling 1 mile, what is the vertical change in feet? (1 mile = 5,280 feet)

#### Solution:
We are given:
- Distance traveled along the slope = 1 mile = 5,280 feet
- Angle of elevation = 2°

We need to find the vertical change (\(x\)), which is the opposite side of the right triangle formed by the slope, the horizontal distance, and the vertical height.

Using trigonometry, specifically the sine function:
\[
\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}
\]
Here, \(\theta = 2^\circ\), the hypotenuse is the distance traveled along the slope (5,280 feet), and the opposite side is the vertical change (\(x\)). So:
\[
\sin(2^\circ) = \frac{x}{5,280}
\]

Solving for \(x\):
\[
x = 5,280 \cdot \sin(2^\circ)
\]

Using a calculator to find \(\sin(2^\circ)\):
\[
\sin(2^\circ) \approx 0.0349
\]

Substituting this value:
\[
x = 5,280 \cdot 0.0349 \approx 184.15 \text{ feet}
\]

Thus, the vertical change is:
\[
\boxed{184.15 \text{ feet}}
\]

---

Final Answers:

1. \(\boxed{106.07 \text{ feet}}\)
2. \(\boxed{30^\circ}\)
3. \(\boxed{19,209 \text{ meters}}\)
4. \(\boxed{184.15 \text{ feet}}\)