← Back to the Law Of Sines And Law Of Cosines worksheet
Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: G-SRT.D.9, G-SRT.D.10, G-SRT.D.11 · Level: Grade 10 Geometry
Problem 1 Area from Two Sides
Area = ½ · 16 · 7 · sin 41° = 56 · 0.6561 = 36.74.
≈ 36.7 ft²
Problem 2 Area with an Obtuse Angle
Area = ½ · 12.5 · 8 · sin 115°. Since sin 115° = sin 65° ≈ 0.9063, area = 50 · 0.9063 = 45.32.
≈ 45.3 ft²
Problem 3 Law of Sines for a Side
bsin 71° = 19sin 44°, so b = 19 sin 71°sin 44° ≈ 17.960.6947 ≈ 25.86.
b ≈ 25.9
Problem 4 Law of Sines for an Angle
sin B19 = sin 72°24, so sin B = 19 · 0.951124 ≈ 0.7529.
B = sin−1(0.7529) ≈ 48.84°. The obtuse choice 131.2° is impossible because 72° + 131.2° is more than 180°.
B ≈ 48.8°
Problem 5 Find the Third Angle First
B = 180° − 53° − 71° = 56°.
asin 53° = 40sin 71°, so a = 40 sin 53°sin 71° ≈ 33.79.
B = 56°, a ≈ 33.8
Problem 6 Law of Cosines for a Side
c2 = 112 + 152 − 2(11)(15) cos 52° = 346 − 330 · 0.6157 = 142.83.
c = √142.83 ≈ 11.95, which rounds to 12.0.
c ≈ 12.0
Problem 7 Obtuse Included Angle
cos 124° = −cos 56° ≈ −0.5592.
c2 = 56.25 + 100 − 150(−0.5592) = 156.25 + 83.88 = 240.13.
c ≈ 15.5.
c ≈ 15.5 in
Problem 8 Largest Angle from Three Sides
The largest angle is opposite 15. cos θ = 7² + 11² − 15²2 · 7 · 11 = 49 + 121 − 225154 = −55154 ≈ −0.3571.
θ = cos−1(−0.3571) ≈ 110.92°.
θ ≈ 110.9°
Problem 9 Choose the Law
(a) Two angles and a side with its opposite angle: Law of Sines.
(b) Two sides and the included angle: Law of Cosines.
(c) Three sides: Law of Cosines.
(d) Two angles and a side with its opposite angle: Law of Sines.
(a) Sines (b) Cosines (c) Cosines (d) Sines
Problem 10 Solve from Three Sides
Largest angle C (opposite 15): cos C = 64 + 121 − 2252 · 8 · 11 = −40176 ≈ −0.2273, so C ≈ 103.14°.
cos A = 121 + 225 − 642 · 11 · 15 = 282330 ≈ 0.8545, so A ≈ 31.29°.
B = 180° − 31.29° − 103.14° ≈ 45.57°.
A ≈ 31.3°, B ≈ 45.6°, C ≈ 103.1°
Problem 11 Solve from Two Sides and an Angle
a2 = 22² + 17² − 2(22)(17) cos 67° = 773 − 748 · 0.3907 = 480.73, so a ≈ 21.93.
sin C = 17 sin 67°21.93 ≈ 0.7137, so C ≈ 45.54° (the side 17 is not the longest, so C is acute).
B = 180° − 67° − 45.54° ≈ 67.46°.
a ≈ 21.9, B ≈ 67.5°, C ≈ 45.5°
Problem 12 Solve from Two Angles and a Side
A = 180° − 39° − 88° = 53°.
b = 31 sin 39°sin 53° ≈ 24.43. c = 31 sin 88°sin 53° ≈ 38.79.
A = 53°, b ≈ 24.4, c ≈ 38.8
Problem 13 Area from Three Sides
Angle C opposite 16: cos C = 81 + 169 − 2562 · 9 · 13 = −6234 ≈ −0.0256, so C ≈ 91.47°.
Area = ½ · 9 · 13 · sin 91.47° ≈ 58.5 · 0.9997 ≈ 58.48.
≈ 58.5 yd²
Problem 14 Area of a Parallelogram
Each triangle has area ½ · 18 · 12.5 · sin 64° = 101.11. Two triangles give 202.23.
≈ 202.2 in²
Problem 15 Distance Across a Lake
AB² = 420² + 365² − 2(420)(365) cos 71° = 309,625 − 306,600 · 0.3256 = 209806.
AB ≈ 458.05.
≈ 458 ft
Problem 16 Two Boats Leave a Dock
The angle between the paths is 112° − 25° = 87°.
d² = 12² + 9² − 2(12)(9) cos 87° = 225 − 216 · 0.0523 = 213.7.
d ≈ 14.62.
≈ 14.6 mi
Problem 17 Observers Tracking a Plane
The angle at the plane is 180° − 41° − 67° = 72°.
(a) The distance is opposite the 67° angle: 3.2 sin 67°sin 72° ≈ 2.950.9511 ≈ 3.10 mi.
(b) Altitude = 3.10 · sin 41° ≈ 2.03 mi.
(a) ≈ 3.10 mi (b) ≈ 2.03 mi
Problem 18 Two Possible Triangles
sin B = 12 sin 30°7 = 67 ≈ 0.8571. So B ≈ 59° or B ≈ 121°; both fit with 30° because each sum is less than 180°.
Triangle 1: C ≈ 91°, c = 7 sin 91°sin 30° ≈ 14.
Triangle 2: C ≈ 29°, c = 7 sin 29°sin 30° ≈ 6.79.
Triangle 1: C ≈ 91°, c ≈ 14. Triangle 2: C ≈ 29°, c ≈ 6.8
Problem 19 Area of a Four-Sided Lot
Triangle 1: ½ · 62 · 48 · sin 75° = 1,488 · 0.9659 = 1437.3.
Triangle 2: ½ · 55 · 40 · sin 110° = 1,100 · 0.9397 = 1033.66.
Total = 2470.96.
≈ 2471 ft²
Problem 20 Drone Return Flight
(a) AC² = 800² + 650² − 2(800)(650) cos 128° = 1,062,500 + 1,040,000 · 0.6157 = 1702788. AC ≈ 1304.91.
(b) sin A = 650 sin 128°1304.91 ≈ 0.3925, so A ≈ 23.11° (acute, since 650 is the shorter side next to the obtuse angle).
(a) ≈ 1304.9 ft (b) ≈ 23.1°