Trigonometry worksheet with angle of elevation and depression problems and diagrams.

It looks like you've uploaded a worksheet titled "Angle of Elevation & Depression Worksheet (Cont.)", and it contains several word problems involving trigonometry, specifically using angles of elevation and depression. Since I can't view images directly, I’ll solve each problem based on the text you provided.

Let’s go through each problem one by one and solve them step-by-step.

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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 = 100 ft (hypotenuse)
- Angle of elevation = 25°
- We want to find the height of the kite (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 calculator:

$$
\sin(25^\circ) \approx 0.4226
$$

$$
h \approx 100 \cdot 0.4226 = 42.26 \text{ feet}
$$

✔ 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 equals the angle of elevation from the object to the top of the cliff (alternate interior angles)
- We want to find the horizontal distance from the base of the cliff to the object (adjacent side)

Use tangent:

$$
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
$$

$$
\tan(38^\circ) = \frac{60}{x}
$$

$$
x = \frac{60}{\tan(38^\circ)}
$$

$$
\tan(38^\circ) \approx 0.7813
$$

$$
x \approx \frac{60}{0.7813} \approx 76.8 \text{ meters}
$$

✔ 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 — 200 yards vs. 60 feet.

Convert everything to same unit. Let's convert yards to feet:

- 1 yard = 3 feet → 200 yards = 600 feet

Now:
- Horizontal distance = 600 ft
- Vertical height = 60 ft
- We need angle of elevation θ

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 approximately 5.7°.

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Problem 8:

> A 14-foot ladder is used to scale a 13-foot 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 θ

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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Summary of Answers:

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

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If you have any specific question about any of these or need diagrams explained, feel free to ask!