Law of Sines Worksheet Answer Key – Precalculus

Problem 1 Area from Two Sides and an Angle

Area = ½ab sin C = ½(14)(9) sin 38° = 63 × 0.6157 ≈ 38.8 square inches.

Problem 2 Heron’s Formula

s = (20 + 37 + 51)/2 = 54. Area = √(54 · 34 · 17 · 3) = √93,636 = 306 square inches.

Problem 3 Law of Sines with AAS

C = 180° − 48° − 71° = 61°.

b = 12 sin 71° ÷ sin 48° ≈ 15.27. c = 12 sin 61° ÷ sin 48° ≈ 14.12.

Problem 4 Law of Cosines for the Third Side

c² = 64 + 121 − 2(8)(11) cos 52° = 185 − 176(0.6157) ≈ 76.64.

c ≈ 8.75.

Problem 5 Third Side from a Figure

Law of Cosines: a² = 49 + 81 − 2(7)(9) cos 64° = 130 − 126(0.4384) ≈ 74.77.

a ≈ 8.65.

Problem 6 Angle from Three Sides

The largest angle is opposite 15: cos C = (64 + 100 − 225)/(2 · 8 · 10) = −61/160 ≈ −0.3813.

C ≈ 112.4°.

Problem 7 Law of Sines with ASA

B = 180° − 36° − 79° = 65°.

a = 20 sin 36° ÷ sin 65° ≈ 12.97. c = 20 sin 79° ÷ sin 65° ≈ 21.66.

Problem 8 SSA with One Triangle

Since a > b, exactly one triangle exists. sin B = 11 sin 50° ÷ 15 ≈ 0.5618, so B ≈ 34.2°.

C = 180° − 50° − 34.2° ≈ 95.8°, and c = 15 sin C ÷ sin 50° ≈ 19.5.

Problem 9 SSA with No Triangle

The height is h = b sin A = 9 sin 40° ≈ 5.79. Since a = 5 is less than h, the side is too short to reach the base line.

Equivalently sin B = 9 sin 40° ÷ 5 ≈ 1.157 > 1. No triangle exists.

Problem 10 SSA with Two Triangles

The height is 13 sin 32° ≈ 6.89, and 6.89 < 9 < 13, so two triangles fit.

sin B = 13 sin 32° ÷ 9 ≈ 0.7654. Triangle 1: B₁ ≈ 49.9°, C₁ ≈ 98.1°, c₁ = 9 sin C₁ ÷ sin 32° ≈ 16.8.

Triangle 2: B₂ = 180° − 49.9° ≈ 130.1°, C₂ ≈ 17.9°, c₂ ≈ 5.2.

Problem 11 Ship on Two Bearings

The interior angle at the turn is 180° − (115° − 40°) = 105°.

d² = 20² + 15² − 2(20)(15) cos 105° = 625 + 155.29 ≈ 780.3.

d ≈ 27.9 miles.

Problem 12 Area from Three Decimal Sides

s = (7.2 + 8.5 + 10.1)/2 = 12.9. Area = √(12.9 · 5.7 · 4.4 · 2.8) = √905.89 ≈ 30.1 square centimeters.

Problem 13 Find the Area After Solving

b = 10 sin 63° ÷ sin 52° ≈ 11.3 and C = 180° − 52° − 63° = 65°.

Area = ½ab sin C = ½(10)(11.31) sin 65° ≈ 51.2 square units.

Problem 14 Area of a Parallelogram

A diagonal splits the parallelogram into two triangles with equal area ½(12)(18) sin 65°. The parallelogram is twice that: 12 · 18 · sin 65° = 216 × 0.9063 ≈ 195.8 square units.

Problem 15 All Three Angles from Three Sides

Largest angle (opposite 20): cos = (144 + 225 − 400)/(2 · 12 · 15) = −31/360, so 94.9°.

Opposite 15: cos = (144 + 400 − 225)/(2 · 12 · 20) = 319/480, so 48.3°.

The last angle is 180° − 94.9° − 48.3° ≈ 36.7° (opposite 12).

Problem 16 Distance Across a Pond

AB² = 210² + 165² − 2(210)(165) cos 74° = 71,325 − 69,300(0.2756) ≈ 52,223.

AB ≈ 228.5 feet.

Problem 17 Locating a Fire from Two Stations

The angle at the fire is 180° − 62° − 48° = 70°.

Law of Sines: distance from A = 8.4 sin 48° ÷ sin 70° ≈ 6.64 miles; distance from B = 8.4 sin 62° ÷ sin 70° ≈ 7.89 miles.

Problem 18 Hiker Back to Camp

The interior angle at the turn is 180° − (140° − 25°) = 65°.

d² = 3.5² + 2.2² − 2(3.5)(2.2) cos 65° ≈ 10.58, so d ≈ 3.3 miles.

sin(angle at camp) = 2.2 sin 65° ÷ 3.253 ≈ 0.6129, so the angle is ≈ 37.8°.

Problem 19 Triangular Lot

s = (120 + 150 + 190)/2 = 230. Area = √(230 · 110 · 80 · 40) = √80,960,000 ≈ 8,998 square feet.

Cost ≈ 8,998 × 0.85 ≈ $7,648.

Problem 20 A Garden with Two Possible Shapes

The height is 18 sin 28° ≈ 8.45 and 8.45 < 11 < 18, so two shapes fit.

sin B = 18 sin 28° ÷ 11 ≈ 0.7682. Shape 1: B ≈ 50.2°, C ≈ 101.8°, AB = 11 sin C ÷ sin 28° ≈ 22.9 ft, area = ½(18)(22.9) sin 28° ≈ 96.9 ft².

Shape 2: B ≈ 129.8°, C ≈ 22.2°, AB ≈ 8.9 ft, area ≈ 37.4 ft².