Angle of Elevation 31 | PDF - Free Printable

Let’s go through each problem one by one. I’ll solve them step by step, check my work, and then give the final answers.

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Problem 1:
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?

We have a right triangle:
- Hypotenuse = 100 ft (the string)
- Angle = 52°
- Opposite side = height of kite (x)

Use sine:
sin(θ) = opposite / hypotenuse
→ sin(52°) = x / 100
→ x = 100 × sin(52°)

Calculate sin(52°) ≈ 0.7880
→ x ≈ 100 × 0.7880 = 78.8

✔ Rounded to nearest tenth: 78.8

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Problem 2:
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?

Angle of depression = angle of elevation from the object to the top of the cliff = 34°

Right triangle:
- Opposite = 40 m (cliff height)
- Adjacent = distance from base (let’s call it d)
- Use tangent: tan(θ) = opposite / adjacent

tan(34°) = 40 / d
→ d = 40 / tan(34°)

tan(34°) ≈ 0.6745
→ d ≈ 40 / 0.6745 ≈ 59.3

✔ Rounded to nearest tenth: 59.3

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Problem 3:
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 the airplane flies over the building.

First, convert units to be consistent.
200 yards = 200 × 3 = 600 feet

Right triangle:
- Opposite = 60 ft (building height)
- Adjacent = 600 ft (distance from takeoff point to building)
- Find angle x° such that tan(x) = opposite / adjacent = 60 / 600 = 0.1

x = arctan(0.1)
arctan(0.1) ≈ 5.71°

✔ Rounded to nearest tenth: 5.7°

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Problem 4:
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)
- Use sine: sin(θ) = opposite / hypotenuse = 13 / 14

sin(θ) ≈ 0.9286
θ = arcsin(0.9286) ≈ 68.2°

✔ Rounded to nearest tenth: 68.2°

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Problem 5:
A person stands at the window of a building so that his eyes are 12.6 m above the level ground. An object is on the ground 58.5 m away from the building on a line directly beneath the person. Compute the angle of depression of the person’s line of sight to the object on the ground.

Angle of depression = angle below horizontal = same as angle of elevation from object to person’s eyes.

Right triangle:
- Opposite = 12.6 m (height)
- Adjacent = 58.5 m (horizontal distance)
- tan(θ) = 12.6 / 58.5 ≈ 0.2154

θ = arctan(0.2154) ≈ 12.15°

✔ Rounded to nearest tenth: 12.2°

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Problem 6:
A ramp is needed to allow vehicles to climb a 2 foot wall. The angle of elevation in order for the vehicles to safely go up must be 30° or less, and the longest ramp available is 5 feet long. Can this ramp be used safely?

Right triangle:
- Opposite = 2 ft (wall height)
- Hypotenuse = 5 ft (ramp length)
- sin(θ) = 2 / 5 = 0.4

θ = arcsin(0.4) ≈ 23.6°

Is 23.6° ≤ 30°? → Yes!

✔ So, YES, the ramp can be used safely.

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Final Answer:
1. 78.8
2. 59.3
3. 5.7°
4. 68.2°
5. 12.2°
6. Yes