Angles of Elevation and Depression Practice - Angle of Elevation ... - Free Printable

Let’s solve each problem one by one, step by step. We’ll use basic trigonometry — sine, cosine, and tangent — based on right triangles.

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Problem 1:
Brian’s kite is flying at the end of 65 m of string. Angle of elevation = 70°. How high is the kite above Brian’s head?

We have:
- Hypotenuse (string) = 65 m
- Angle = 70°
- Opposite side = height (what we’re solving for)

Use sine:
sin(θ) = opposite / hypotenuse
→ sin(70°) = x / 65
→ x = 65 × sin(70°)

Calculate sin(70°): ≈ 0.9397
→ x ≈ 65 × 0.9397 ≈ 61.08 m

✔ Final Answer for #1: 61.08 meters

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Problem 2:
Airplane altitude = 1200 m. Angle of depression to building = 28°. Find distance from plane to building.

Angle of depression = angle below horizontal → same as angle inside triangle at top.

So in the right triangle:
- Opposite side = 1200 m (altitude)
- Hypotenuse = distance from plane to building (x)
- Angle = 28°

Use sine:
sin(28°) = opposite / hypotenuse = 1200 / x
→ x = 1200 / sin(28°)

sin(28°) ≈ 0.4695
→ x ≈ 1200 / 0.4695 ≈ 2556.07 m

✔ Final Answer for #2: 2556.07 meters

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Problem 3:
From ground, 12 ft from flagpole base. Angle of elevation to top = 53°. How tall is flagpole?

Adjacent side = 12 ft
Opposite side = height (o)
Angle = 53°

Use tangent:
tan(53°) = opposite / adjacent = o / 12
→ o = 12 × tan(53°)

tan(53°) ≈ 1.3270
→ o ≈ 12 × 1.3270 ≈ 15.92 ft

✔ Final Answer for #3: 15.92 feet

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Problem 4:
Plane at 265 m above sea level. Angles of depression to two ships: 35° and 25°. Both ships due east. How far apart are they?

This means we have two right triangles sharing the same vertical side (265 m).

For first ship (35° depression):
tan(35°) = 265 / a₁ → a₁ = 265 / tan(35°)
tan(35°) ≈ 0.7002 → a₁ ≈ 265 / 0.7002 ≈ 378.46 m

For second ship (25° depression):
tan(25°) = 265 / a₂ → a₂ = 265 / tan(25°)
tan(25°) ≈ 0.4663 → a₂ ≈ 265 / 0.4663 ≈ 568.24 m

Distance between ships = a₂ - a₁ = 568.24 - 378.46 = 189.78 m

Wait — let me double-check subtraction:

568.24
-378.46
= 189.78 m

But in the original work, it says “184.83m” — that might be a calculation error. Let me recalculate with more precision.

Actually, let's use calculator values:

tan(35°) = 0.7002075382
a₁ = 265 / 0.7002075382 ≈ 378.462

tan(25°) = 0.46630765815
a₂ = 265 / 0.46630765815 ≈ 568.294

Difference: 568.294 - 378.462 = 189.832 m

Hmm — still around 189.83 m. But the handwritten answer says 184.83m — maybe they used different rounding? Or perhaps misread angles?

Wait — maybe the angles are measured differently? No — standard interpretation is correct.

Alternatively, perhaps the problem meant angles from vertical? Unlikely — angle of depression is always from horizontal.

I think the correct answer is approximately 189.83 m, but since the student wrote 184.83, let’s check if they swapped something.

Wait — what if they did tan(35) = a/265? That would be wrong — because tan(angle) = opposite/adjacent = 265/a, not a/265.

If someone mistakenly did a = 265 * tan(35), then:

a₁ = 265 * tan(35) ≈ 265 * 0.7002 ≈ 185.55
a₂ = 265 * tan(25) ≈ 265 * 0.4663 ≈ 123.57
Then difference = 185.55 - 123.57 = 61.98 — doesn’t match.

Alternatively, if they computed distances along ground correctly but subtracted wrong?

Wait — another possibility: maybe they calculated the distance between the points directly under the plane and each ship, which is what we did — so 189.83 m is correct.

But looking back at the image, the student has:

“tan 35 = 265 / a → a = 265 / tan35 ≈ 378.46”

“tan 25 = 265 / a → a = 265 / tan25 ≈ 568.24”

Then “568.24 - 378.46 = 189.78” — but they wrote “184.83m” in box — probably typo or arithmetic mistake.

So correct value is ≈189.78 m

But to match common textbook answers, sometimes they round early.

Let me compute exactly:

Using precise values:

tan(35°) = tan(35) = 0.7002075382097097
a1 = 265 / 0.7002075382097097 = 378.462...

tan(25°) = 0.4663076581549986
a2 = 265 / 0.4663076581549986 = 568.294...

Difference: 568.294 - 378.462 = 189.832

Rounded to two decimals: 189.83 m

But since the student got 184.83, and I see no reason for that, I’ll go with correct math.

However — wait! In the diagram, the student labeled the 35° triangle with “La)” and 25° with “(a)” — maybe they assigned labels backwards? If 35° corresponds to farther ship? No — smaller angle means farther away.

Actually, 25° < 35°, so ship with 25° depression is farther — yes, so a2 > a1 — correct.

Perhaps the problem meant angles of elevation from ships? Same thing.

I think the correct answer is 189.83 m, but let’s note that the student may have made an arithmetic error.

To be safe, I’ll report the accurate value.

✔ Final Answer for #4: 189.83 meters

*(Note: The handwritten answer of 184.83 appears to be incorrect; likely a calculation slip.)*

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Problem 5:
Kite string = 100 ft. Angle of elevation = 52°. Find:
(a) Height off ground
(b) Horizontal distance from man to spot under kite

Right triangle:
Hypotenuse = 100 ft
Angle = 52°

(a) Height = opposite side → use sine
sin(52°) = h / 100 → h = 100 × sin(52°)
sin(52°) ≈ 0.7880 → h ≈ 78.80 ft

(b) Horizontal distance = adjacent side → use cosine
cos(52°) = d / 100 → d = 100 × cos(52°)
cos(52°) ≈ 0.6157 → d ≈ 61.57 ft

✔ Final Answers for #5:
Height = 78.80 feet
Horizontal distance = 61.57 feet

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Problem 6:
Cliff height = 40 m. Angle of depression to object = 34°. Object is level with base of cliff. How far is object from base of cliff?

Angle of depression = 34° → same as angle at top of cliff inside triangle.

So:
Opposite side = 40 m (cliff height)
Adjacent side = distance from base to object (x)
Angle = 34°

Use tangent:
tan(34°) = opposite / adjacent = 40 / x
→ x = 40 / tan(34°)

tan(34°) ≈ 0.6745
→ x ≈ 40 / 0.6745 ≈ 59.30 m

✔ Final Answer for #6: 59.30 meters

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Now compiling all final answers:

Final Answer:
1. 61.08 meters
2. 2556.07 meters
3. 15.92 feet
4. 189.83 meters
5. Height: 78.80 feet; Horizontal distance: 61.57 feet
6. 59.30 meters