Right Triangle Trigonometry Worksheet Answer Key – Geometry

Problem 1 Naming the Ratios

Opposite A is BC = 40, adjacent is AC = 9, hypotenuse is AB = 41.

sin A = 4041, cos A = 941, tan A = 409.

sin A = 40/41, cos A = 9/41, tan A = 40/9

Problem 2 Ratios for the Other Angle

From angle B, the opposite side is AC = 8 and the adjacent side is BC = 15.

sin B = 817, cos B = 1517, tan B = 815.

sin B = 8/17, cos B = 15/17, tan B = 8/15

Problem 3 Complementary Angles

Sine and cosine of complementary angles are equal.

(a) cos 67° = sin(90° − 67°) = sin 23°.

(b) x = 90 − 28 = 62.

(c) 24° + 66° = 90°, so cos 66° = sin 24° ≈ 0.4067.

(a) sin 23° (b) x = 62 (c) 0.4067

Problem 4 Missing Leg with Tangent

Opposite is x, adjacent is 14, so tan 35° = x14.

x = 14 · tan 35° ≈ 14 · 0.7002 = 9.8.

x ≈ 9.8

Problem 5 Hypotenuse with Cosine

Adjacent is 11 and the hypotenuse is x: cos 52° = 11x.

x = 11cos 52° ≈ 110.6157 ≈ 17.87.

x ≈ 17.9

Problem 6 Leg with Sine

BC is opposite angle A. sin 41° = BC23, so BC = 23 · sin 41° ≈ 23 · 0.6561 = 15.09.

BC ≈ 15.1

Problem 7 Angle from Two Legs

tan θ = 712, so θ = tan−1(7/12) ≈ 30.26°.

θ ≈ 30.3°

Problem 8 Angle from a Sine Ratio

sin θ = 1319 ≈ 0.6842, so θ = sin−1(0.6842) ≈ 43.17°.

θ ≈ 43.2°

Problem 9 Isosceles Right Triangle

The sides are in the ratio 1 : 1 : √2, so the hypotenuse is 14√2.

14 · 1.4142 ≈ 19.8.

14√2 cm ≈ 19.8 cm

Problem 10 Hypotenuse and Short Leg

The short leg is half the hypotenuse: 22 ÷ 2 = 11. The long leg is short leg · √3 = 11√3 ≈ 19.05.

Short leg 11 ft; long leg 11√3 ft ≈ 19.1 ft

Problem 11 Starting from the Long Leg

Long leg = short leg · √3, so short leg BC = 12√3 ÷ √3 = 12. Hypotenuse = 2 · 12 = 24.

BC = 12 in, AB = 24 in

Problem 12 Solving from Two Legs

AB = √(20² + 21²) = √841 = 29.

tan A = 2120, so A = 46.4° ≈ 46.4°. Then B = 90° − 46.4° = 43.6° (check: tan⁻¹(20/21) ≈ 43.6°).

AB = 29, A ≈ 46.4°, B ≈ 43.6°

Problem 13 Angle and Hypotenuse

B = 90° − 37° = 53°.

BC = 26 · sin 37° ≈ 26 · 0.6018 = 15.65. AC = 26 · cos 37° ≈ 26 · 0.7986 = 20.76.

B = 53°, BC ≈ 15.6, AC ≈ 20.8

Problem 14 Angle and One Leg

A = 90° − 62° = 28°. From angle B, BC is adjacent and AC is opposite.

AC = 9.5 · tan 62° ≈ 9.5 · 1.8807 = 17.87. AB = 9.5cos 62° ≈ 20.24.

A = 28°, AC ≈ 17.9 m, AB ≈ 20.2 m

Problem 15 Surveying a Water Tower

The part above eye level: 120 · tan 34° ≈ 120 · 0.6745 = 80.94 ft. Add the eye height: 80.94 + 5.5 = 86.44.

≈ 86.4 ft

Problem 16 Boat Below the Cliff

The angle at the boat equals 18° (alternate interior angles). The 210 ft side is opposite, the distance d is adjacent: tan 18° = 210d.

d = 210tan 18° ≈ 646.31.

≈ 646 ft

Problem 17 Wheelchair Ramp

Rise is opposite the angle. Ramp length (hypotenuse): 2.5sin 4.8° ≈ 29.88 ft. Horizontal distance (adjacent): 2.5tan 4.8° ≈ 29.77 ft.

Ramp ≈ 29.9 ft long, covering ≈ 29.8 ft horizontally

Problem 18 Ladder and Window

(a) Height = 16 · sin 74° ≈ 16 · 0.9613 = 15.38 ft.

(b) Base distance = 16 · cos 74° ≈ 16 · 0.2756 = 4.41 ft.

(c) 15.38 is less than 15.5, so the ladder falls short of the sill.

(a) ≈ 15.4 ft (b) ≈ 4.4 ft (c) No

Problem 19 Two Sightings of a Tower

Let h be the height and d the first distance. Then h = d · tan 25° and h = (d − 90) · tan 40°.

d · tan 25° = (d − 90) · tan 40° gives d(tan 40° − tan 25°) = 90 · tan 40°.

tan 40° ≈ 0.8391, tan 25° ≈ 0.4663, so d ≈ 75.52 ÷ 0.3728 ≈ 202.6. Then h ≈ 202.6 · 0.4663 ≈ 94.5.

≈ 94 ft

Problem 20 Roof Rafters

Each side covers half of the span: 15 ft horizontally.

(a) Height = 15 · tan 33° ≈ 15 · 0.6494 = 9.74 ft.

(b) Roof length = 15cos 33° ≈ 150.8387 ≈ 17.89 ft. Add the overhang: 17.89 + 1.5 = 19.39 ft.

(a) ≈ 9.7 ft (b) ≈ 19.4 ft