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Trigonometry in the Right Triangle · Ages 15–16

Finding Angles from Two Sides in Right Triangles

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For 10th grade geometry students. For each right triangle, identify the correct ratio, calculate the unknown angle to the nearest tenth of a degree, and find the missing acute angle.

  1. Opposite side = 3, Hypotenuse = 5. Find the angle opposite the side of length 3.

  2. Adjacent side = 8, Hypotenuse = 10. Find the angle adjacent to the side of length 8.

  3. Opposite side = 4, Adjacent side = 4. Find the angle opposite the first leg.

  4. Opposite side = 5, Hypotenuse = 13. Find the angle opposite the side of length 5.

  5. Opposite side = 7, Adjacent side = 12. Find the angle opposite the side of length 7.

  6. In right triangle ABC, angle C is 90°, side AC (adjacent to A) = 9, and hypotenuse AB = 15. Find angle A and angle B.

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Finding Angles from Two Sides in Right Triangles

Answer key

  1. Opposite side = 3, Hypotenuse = 5. Find the angle opposite the side of length 3.

    sin(θ) = 3/5 = 0.6. θ = sin⁻¹(0.6) ≈ 36.9°.

  2. Adjacent side = 8, Hypotenuse = 10. Find the angle adjacent to the side of length 8.

    cos(θ) = 8/10 = 0.8. θ = cos⁻¹(0.8) ≈ 36.9°.

  3. Opposite side = 4, Adjacent side = 4. Find the angle opposite the first leg.

    tan(θ) = 4/4 = 1. θ = tan⁻¹(1) = 45.0°.

  4. Opposite side = 5, Hypotenuse = 13. Find the angle opposite the side of length 5.

    sin(θ) = 5/13 ≈ 0.3846. θ = sin⁻¹(0.3846) ≈ 22.6°.

  5. Opposite side = 7, Adjacent side = 12. Find the angle opposite the side of length 7.

    tan(θ) = 7/12 ≈ 0.5833. θ = tan⁻¹(0.5833) ≈ 30.3°.

  6. In right triangle ABC, angle C is 90°, side AC (adjacent to A) = 9, and hypotenuse AB = 15. Find angle A and angle B.

    cos(A) = 9/15 = 0.6, so A = cos⁻¹(0.6) ≈ 53.1°. Angle B = 90° - 53.1° = 36.9°.