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

Let's solve each problem step by step using trigonometry, specifically the sine, cosine, and tangent functions. We'll use right triangle relationships involving angle of elevation and angle of depression.

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Problem 5:

> A man flies a kite with a 100-foot string. The angle of elevation of the string is 52°. How high off the ground is the kite?

#### Diagram:
- Hypotenuse = 100 ft (length of string)
- Angle of elevation = 52°
- We need to find the opposite side (height of kite)

Use sine:
$$
\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}
$$
$$
\sin(52^\circ) = \frac{h}{100}
$$
$$
h = 100 \cdot \sin(52^\circ)
$$

Using calculator:
$$
\sin(52^\circ) \approx 0.7880
$$
$$
h \approx 100 \cdot 0.7880 = 78.8 \text{ feet}
$$

✔ Answer: The kite is 78.8 feet high.

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Problem 6:

> From the top of a vertical cliff 40 m high, the angle of depression of an object that is level with the base of the cliff is 34°. How far is the object from the base of the cliff?

#### Understanding:
- Height of cliff = 40 m (vertical leg)
- Angle of depression = 34° → this is the angle between the horizontal line from the top and the line of sight to the object.
- Since the object is at the base, the horizontal distance is what we're solving for.

Angle of depression = angle of elevation from the object up to the observer → so we have a right triangle where:
- Opposite = 40 m
- Angle = 34°
- Adjacent = distance from base (let’s call it $ x $)

Use tangent:
$$
\tan(34^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{40}{x}
$$
$$
x = \frac{40}{\tan(34^\circ)}
$$

$$
\tan(34^\circ) \approx 0.6745
$$
$$
x \approx \frac{40}{0.6745} \approx 59.3 \text{ meters}
$$

✔ Answer: The object is 59.3 meters from the base of the cliff.

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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 the airplane flies in a straight line and the angle of elevation remains constant until it flies over the building.

#### Note:
Units are mixed — 200 yards and 60 feet. Convert to same unit.

We’ll convert yards to feet:
$$
200 \text{ yards} = 200 \times 3 = 600 \text{ feet}
$$

Now:
- Opposite = height of building = 60 ft
- Adjacent = horizontal distance = 600 ft
- Find angle of elevation $ x^\circ $

Use tangent:
$$
\tan(x) = \frac{60}{600} = 0.1
$$
$$
x = \tan^{-1}(0.1) \approx 5.7^\circ
$$

✔ Answer: The plane must take off at an angle of 5.7°.

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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?

#### Right triangle:
- Hypotenuse = 14 ft (ladder)
- Opposite = 13 ft (wall height)
- Find angle of elevation $ x^\circ $

Use sine:
$$
\sin(x) = \frac{13}{14}
$$
$$
x = \sin^{-1}\left(\frac{13}{14}\right)
$$
$$
\frac{13}{14} \approx 0.9286
$$
$$
x \approx \sin^{-1}(0.9286) \approx 68.2^\circ
$$

✔ Answer: The ladder must be placed at an angle of 68.2°.

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✔ Final Answers:

| Problem | Answer |
|--------|--------|
| 5 | 78.8 feet |
| 6 | 59.3 meters |
| 7 | 5.7° |
| 8 | 68.2° |

All values rounded to the nearest tenth as requested.

Let me know if you'd like diagrams or further explanation!