Right Triangle Trigonometry Practice Worksheet - Finding Missing Sides and Angles

Let’s solve each problem one by one. We’ll use right triangle trigonometry: sine, cosine, and tangent. Remember:

- sin(θ) = opposite / hypotenuse
- cos(θ) = adjacent / hypotenuse
- tan(θ) = opposite / adjacent

We’ll also use the Pythagorean theorem when needed:
a² + b² = c² (for right triangles, where c is the hypotenuse)

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Problem 1: Find angle θ at A in triangle ABC (right angle at C), sides: AC=12, AB=13

This is a right triangle with right angle at C. So side AB is the hypotenuse (13). Side AC is adjacent to angle θ (at A). So we can use cosine.

cos(θ) = adjacent / hypotenuse = AC / AB = 12 / 13

θ = cos⁻¹(12/13) ≈ cos⁻¹(0.9231) ≈ 22.6°

✔ Check: sin(θ) should be BC/AB. Let’s find BC using Pythagoras:
BC = √(13² - 12²) = √(169 - 144) = √25 = 5 → sin(θ)=5/13≈0.3846 → θ≈22.6° ✔️

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Problem 2: Find angle θ at A in triangle ABC (right angle at B), sides: AB=13, BC=4

Right angle at B → so AC is hypotenuse? Wait — let’s label carefully.

Triangle ABC, right angle at B → so legs are AB and BC, hypotenuse is AC.

But we’re given AB=13, BC=4. Angle θ is at A.

So for angle θ at A:
- Opposite side = BC = 4
- Adjacent side = AB = 13
→ tan(θ) = opposite / adjacent = 4 / 13

θ = tan⁻¹(4/13) ≈ tan⁻¹(0.3077) ≈ 17.1°

✔ Check: Hypotenuse AC = √(13² + 4²) = √(169+16)=√185≈13.6 → sin(θ)=4/13.6≈0.294 → θ≈17.1° ✔️

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Problem 3: Find angle θ at A in triangle ABC (right angle at C), sides: AC=6, AB=9

Wait — if right angle is at C, then AB is hypotenuse? But AB=9, AC=6 → then BC = √(81 - 36) = √45 ≈ 6.7

Angle θ at A → adjacent = AC = 6, hypotenuse = AB = 9

cos(θ) = 6/9 = 2/3 ≈ 0.6667 → θ = cos⁻¹(2/3) ≈ 48.2°

✔ Check: sin(θ) = BC/AB = √45 / 9 ≈ 6.708/9 ≈ 0.745 → θ≈48.2° ✔️

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Problem 4: Find angle θ at A in triangle ABC (right angle at C), sides: BC=10, AC=11.8

Right angle at C → legs AC and BC, hypotenuse AB.

Angle θ at A → opposite = BC = 10, adjacent = AC = 11.8

tan(θ) = 10 / 11.8 ≈ 0.8475 → θ = tan⁻¹(0.8475) ≈ 40.3°

✔ Check: hypotenuse AB = √(10² + 11.8²) = √(100 + 139.24) = √239.24 ≈ 15.47 → sin(θ)=10/15.47≈0.646 → θ≈40.3° ✔️

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Problem 5: Find angle θ at B in triangle ABC (right angle at C), sides: AC=7.7, AB=14

Right angle at C → AB is hypotenuse (14), AC is opposite to angle θ at B.

Wait — angle θ is at B. So:

- For angle at B:
- Opposite side = AC = 7.7
- Hypotenuse = AB = 14

sin(θ) = 7.7 / 14 ≈ 0.55 → θ = sin⁻¹(0.55) ≈ 33.4°

✔ Check: adjacent = BC = √(14² - 7.7²) = √(196 - 59.29) = √136.71 ≈ 11.69 → cos(θ)=11.69/14≈0.835 → θ≈33.4° ✔️

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Problem 6: Find angle θ at B in triangle ABC (right angle at C), sides: BC=4, AB=5

Right angle at C → AB is hypotenuse (5), BC is adjacent to angle θ at B.

Wait — angle θ at B:

- Adjacent = BC = 4
- Hypotenuse = AB = 5

cos(θ) = 4/5 = 0.8 → θ = cos⁻¹(0.8) ≈ 36.9°

✔ Check: opposite = AC = √(25 - 16) = √9 = 3 → sin(θ)=3/5=0.6 → θ≈36.9° ✔️

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Problem 7: Find angle θ at B in triangle ABC (right angle at C), sides: AB=11, BC=4.4

Right angle at C → AB is hypotenuse (11), BC is adjacent to angle θ at B.

cos(θ) = adjacent/hypotenuse = BC / AB = 4.4 / 11 = 0.4

θ = cos⁻¹(0.4) ≈ 66.4°

✔ Check: opposite = AC = √(121 - 19.36) = √101.64 ≈ 10.08 → sin(θ)=10.08/11≈0.916 → θ≈66.4° ✔️

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Problem 8: Find angle θ at B in triangle ABC (right angle at C), sides: BC=3, AC=3

Right angle at C → legs BC and AC both 3 → it’s an isosceles right triangle!

Angle at B: opposite = AC = 3, adjacent = BC = 3 → tan(θ)=3/3=1 → θ=45.0°

✔ Obvious check: angles in triangle sum to 180°, right angle at C → other two angles equal → 45° each ✔️

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Now Part 2: Find missing sides.

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Problem 9: Find side x (hypotenuse) in triangle ABC (right angle at C), angle at B=37°, side BC=11

Angle at B = 37°, side BC = 11 → this is adjacent to angle B.

We want hypotenuse AB = x.

cos(37°) = adjacent / hypotenuse = 11 / x

→ x = 11 / cos(37°)

cos(37°) ≈ 0.7986 → x ≈ 11 / 0.7986 ≈ 13.8

✔ Check: opposite = AC = x * sin(37°) ≈ 13.8 * 0.6018 ≈ 8.3 → Pythagoras: 11² + 8.3² ≈ 121 + 68.89 = 189.89 → √189.89≈13.78≈13.8 ✔️

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Problem 10: Find side x (opposite to 36°) in triangle ABC (right angle at C), angle at A=36°, side AC=13

Angle at A = 36°, side AC = 13 → this is adjacent to angle A.

We want opposite side BC = x.

tan(36°) = opposite / adjacent = x / 13

→ x = 13 * tan(36°)

tan(36°) ≈ 0.7265 → x ≈ 13 * 0.7265 ≈ 9.4

✔ Check: hypotenuse AB = 13 / cos(36°) ≈ 13 / 0.8090 ≈ 16.07 → sin(36°)=x/16.07≈0.5878 → x≈9.4 ✔️

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Problem 11: Find side x (adjacent to 50.1°) in triangle ABC (right angle at C), angle at A=50.1°, side AC=5

Wait — diagram shows: right angle at C, angle at A=50.1°, side AC=5 → that’s adjacent to angle A.

We want side BC = x → which is opposite to angle A.

tan(50.1°) = opposite / adjacent = x / 5

→ x = 5 * tan(50.1°)

tan(50.1°) ≈ 1.196 → x ≈ 5 * 1.196 ≈ 6.0

✔ Check: hypotenuse AB = 5 / cos(50.1°) ≈ 5 / 0.6428 ≈ 7.78 → sin(50.1°)=x/7.78≈0.767 → x≈6.0 ✔️

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Problem 12: Find side x (opposite to 60°) in triangle ABC (right angle at C), angle at B=60°, side AB=11

Angle at B = 60°, hypotenuse AB = 11.

Side x is opposite to angle B → so sin(60°) = x / 11

→ x = 11 * sin(60°) = 11 * (√3/2) ≈ 11 * 0.8660 ≈ 9.5

✔ Check: adjacent = BC = 11 * cos(60°) = 11 * 0.5 = 5.5 → Pythagoras: 5.5² + 9.5² = 30.25 + 90.25 = 120.5 → √120.5≈10.98≈11 ✔️

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Final Answers:

1) 22.6°
2) 17.1°
3) 48.2°
4) 40.3°
5) 33.4°
6) 36.9°
7) 66.4°
8) 45.0°
9) 13.8
10) 9.4
11) 6.0
12) 9.5

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Final Answer:
1) 22.6°
2) 17.1°
3) 48.2°
4) 40.3°
5) 33.4°
6) 36.9°
7) 66.4°
8) 45.0°
9) 13.8
10) 9.4
11) 6.0
12) 9.5