Geometry Law of Sines and Law of Cosines Worksheet Solutions (PDF)

Geometry math worksheet solutions: Law of Sines and Law of Cosines – Maths PDF

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This law of sines cosines answers for Geometry covers law of sines and law of cosines (Common Core G-SRT.D.9, G-SRT.D.10, G-SRT.D.11), with a free PDF to print.

Use this answer key to check the 20 problems on law of sines and cosines. Every solution shows the formula or theorem used, each step of the work, and the final answer in bold. If an answer differs from yours, compare the steps line by line to find where the work changed.

Geometry · Common Core State Standards: G-SRT.D.9, G-SRT.D.10, G-SRT.D.11

Law of sines cosines answers: exercises 1 to 10

Exercise 1 – Area from SAS

$\frac{1}{2} \cdot 8 \cdot 11 \cdot \text{sin}\,30^{\circ} = 44 \cdot 0.5 = 22$

22 square units

Exercise 2 – Area of a sail

$\frac{1}{2} \cdot 7.5 \cdot 12 \cdot \text{sin}\,58^{\circ} = 45 \cdot 0.8480 = 38.2$

38.2 square meters

Exercise 3 – Law of Sines: find a side

b is about 24.9.

$\frac{b}{\text{sin}\,72^{\circ}} = \frac{15}{\text{sin}\,35^{\circ}}$

$b = \frac{15 \cdot 0.9511}{0.5736} = 24.9$

b = 24.9

Exercise 4 – Law of Sines: find an angle

$\text{sin}\,B = \frac{15 \cdot \text{sin}\,50^{\circ}}{18} = 0.6384$

$B = \text{sin}^{-1}(0.6384) = 39.7^{\circ}$

B = 39.7°

Exercise 5 – ASA

C = 67 degrees.

$C = 180^{\circ} – 52^{\circ} – 61^{\circ} = 67^{\circ}$.

$\frac{a}{\text{sin}\,52^{\circ}} = \frac{20}{\text{sin}\,67^{\circ}}$, so $a = 17.1$.

C = 67°; a = 17.1

Exercise 6 – Law of Cosines: SAS

c is about 11.3.

$c^{2} = 12^{2} + 15^{2} – 2 \cdot 12 \cdot 15 \cdot \text{cos}\,48^{\circ}$

$c^{2} = 369 – 360 \cdot 0.6691 = 128.1$

$c = 11.3$

c = 11.3

Exercise 7 – Law of Cosines: SSS

$5^{2} = 7^{2} + 9^{2} – 2 \cdot 7 \cdot 9 \cdot \text{cos}\,A$

$25 = 130 – 126 \text{cos}\,A$

$\text{cos}\,A = \frac{105}{126} = 0.8333$

$A = \text{cos}^{-1}(0.8333) = 33.6^{\circ}$

A = 33.6°

Exercise 8 – Obtuse angle

The largest angle is opposite 8.

$8^{2} = 4^{2} + 5^{2} – 2 \cdot 4 \cdot 5 \cdot \text{cos}\,C$

$\text{cos}\,C = \frac{16 + 25 – 64}{40} = -0.575$

$C = \text{cos}^{-1}(-0.575) = 125.1^{\circ}$

125.1°

Exercise 9 – Right triangle check

$25^{2} = 7^{2} + 24^{2} – 2 \cdot 7 \cdot 24 \cdot \text{cos}\,C$

$625 = 625 – 336 \text{cos}\,C$

$\text{cos}\,C = 0$, so $C = 90^{\circ}$.

Right angle opposite the side 25

Exercise 10 – Which law?

(a) Law of Sines. (b) Law of Cosines. (c) Law of Cosines. (d) Law of Sines.

a: sines; b: cosines; c: cosines; d: sines

Exercises 11 and up

Exercise 11 – Solve a triangle (SAS)

Triangle ABC with c about 7.8.

$c^{2} = 100 + 144 – 240 \text{cos}\,40^{\circ} = 60.1$, so $c = 7.8$.

$\text{sin}\,A = \frac{10 \cdot \text{sin}\,40^{\circ}}{7.8} = 0.8288$, so $A = 56.0^{\circ}$.

$B = 180^{\circ} – 40^{\circ} – 56.0^{\circ} = 84.0^{\circ}$.

c = 7.8; A = 56.0°; B = 84.0°

Exercise 12 – Angle opposite the shorter side

$\text{sin}\,B = \frac{18 \cdot \text{sin}\,58^{\circ}}{22} = 0.6939$

$B = \text{sin}^{-1}(0.6938) = 43.9^{\circ}$

B = 43.9°

Exercise 13 – Equilateral triangle area

$\frac{1}{2} \cdot 10 \cdot 10 \cdot \text{sin}\,60^{\circ} = 50 \cdot \frac{\sqrt{3}}{2} = 25\sqrt{3}$.

25√3, about 43.3

Exercise 14 – Angle from area

$30 = \frac{1}{2} \cdot 10 \cdot 12 \cdot \text{sin}\,C$

$\text{sin}\,C = \frac{30}{60} = 0.5$

$C = 30^{\circ}$

30°

Exercise 15 – Across a lake

$AB^{2} = 420^{2} + 380^{2} – 2 \cdot 420 \cdot 380 \cdot \text{cos}\,48^{\circ}$

$AB^{2} = 320{,}800 – 319{,}200 \cdot 0.6691 = 107,214$

$AB = 327$

About 327 feet

Exercise 16 – Ship's return

$d^{2} = 40^{2} + 55^{2} – 2 \cdot 40 \cdot 55 \cdot \text{cos}\,110^{\circ}$

$d^{2} = 4625 + 4400 \cdot 0.3420 = 6,130$

$d = 78$

About 78 miles

Exercise 17 – Fire lookouts

Lookouts at A and B, fire at C.

Angle at the fire: $180^{\circ} – 62^{\circ} – 71^{\circ} = 47^{\circ}$.

The side opposite the $71^{\circ}$ angle is the distance from the 62° lookout.

$\frac{d}{\text{sin}\,71^{\circ}} = \frac{800}{\text{sin}\,47^{\circ}}$, so $d = 1034$

About 1034 feet

Exercise 18 – Hikers apart

$d^{2} = 5.4^{2} + 3.8^{2} – 2 \cdot 5.4 \cdot 3.8 \cdot \text{cos}\,112^{\circ}$

$d^{2} = 43.6 + 41.04 \cdot 0.3746 = 59.0$

$d = 7.7$

About 7.7 km

Exercise 19 – Garden lot cost

Area: $\frac{1}{2} \cdot 120 \cdot 150 \cdot \text{sin}\,35^{\circ} = 9000 \cdot 0.5736 = 5,162$ square feet.

Cost: $\$0.45 \cdot 5,162 = \$2,323$.

Area about 5,162 sq ft; cost about $2,323

Exercise 20 – Order the sides

The longest side is opposite the largest angle, 80°.

$\frac{b}{\text{sin}\,65^{\circ}} = \frac{6}{\text{sin}\,35^{\circ}}$, so $b = 9.5$.

$c = \frac{6 \cdot \text{sin}\,80^{\circ}}{\text{sin}\,35^{\circ}} = 10.3$.

Longest side c = 10.3; b = 9.5

Key facts behind this law of sines cosines answers

Key facts

Area = (1/2) a b sin C.
Law of Sines: a/sin A = b/sin B = c/sin C.
Law of Cosines: c squared = a squared + b squared – 2ab cos C.
AAS or ASA: use sines. SAS or SSS: use cosines.
When the included angle is 90 degrees the Law of Cosines becomes the Pythagorean theorem.

How to use this law of sines cosines answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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