Solved) - Name Date Angle of Elevation & Depression Worksheet ... - Free Printable

Let's solve each of the problems on the worksheet step by step. Since I can't see the image directly, I will interpret and solve the problems based on your description and the visible text.

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Problem 5:

A man flies a kite with a 100-foot string. The angle of elevation of the string is 25°. How high off the ground is the kite?

#### Solution:
We are given:
- Length of string (hypotenuse) = 100 ft
- Angle of elevation = 25°
- We need to find the height (opposite side)

Use sine function:
$$
\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}
$$
$$
\sin(25^\circ) = \frac{h}{100}
$$
$$
h = 100 \cdot \sin(25^\circ)
$$

Using a calculator:
$$
\sin(25^\circ) \approx 0.4226
$$
$$
h \approx 100 \cdot 0.4226 = 42.26 \text{ ft}
$$

✔ Answer: The kite is approximately 42.3 feet high.

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Problem 6:

From the top of a vertical cliff 60 m high, the angle of depression of an object is level with the base of the cliff is 38°. How far is the object from the base of the cliff?

#### Solution:
- Height of cliff = 60 m (this is the opposite side)
- Angle of depression = 38° → this is equal to the angle of elevation from the object to the top of the cliff
- We want to find the horizontal distance (adjacent side)

Use tangent:
$$
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
$$
$$
\tan(38^\circ) = \frac{60}{x}
$$
$$
x = \frac{60}{\tan(38^\circ)}
$$

$$
\tan(38^\circ) \approx 0.7813
$$
$$
x \approx \frac{60}{0.7813} \approx 76.8 \text{ m}
$$

✔ Answer: The object is approximately 76.8 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 that the airplane flies in a straight line and the angle of elevation remains constant until the airplane flies over the building.

#### Note: Units are mixed — convert to same unit.

Convert 200 yards to feet:
$$
200 \text{ yd} = 200 \times 3 = 600 \text{ ft}
$$

Now:
- Opposite side = height of building = 60 ft
- Adjacent side = horizontal distance = 600 ft
- Find angle of elevation $ \theta $

Use tangent:
$$
\tan(\theta) = \frac{60}{600} = 0.1
$$
$$
\theta = \tan^{-1}(0.1) \approx 5.71^\circ
$$

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

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Problem 8:

A 14 ft ladder is used to scale a 13 ft wall. What angle of elevation must the ladder be situated in order to reach the top of the wall?

#### Solution:
- Ladder = hypotenuse = 14 ft
- Wall height = opposite side = 13 ft
- Find angle of elevation $ \theta $

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

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

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 5 | ≈ 42.3 feet |
| 6 | ≈ 76.8 meters |
| 7 | ≈ 5.7° |
| 8 | ≈ 68.2° |

Let me know if you'd like diagrams or explanations for any specific problem!