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

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

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This laws sines cosines answers for Precalculus covers law of sines and law of cosines (Common Core G-SRT.D.10, G-SRT.D.11, G-SRT.C.8), with a free PDF to print.

Use this key to check each triangle. Solutions show which law is used, the setup and the rounded result, with final answers in bold.

Precalculus · Common Core State Standards: G-SRT.D.10, G-SRT.D.11, G-SRT.C.8

Laws sines cosines answers: exercises 1 to 10

Exercise 1 – Right triangle: find a leg

\(\sin35^{\circ}=\frac{x}{20}\), so \(x=20\sin35^{\circ}\approx 11.47\).

\(11.47\)

Exercise 2 – Right triangle: tangent

\(\tan62^{\circ}=\frac{h}{8}\), so \(h=8\tan62^{\circ}\approx 15.05\).

\(15.05\text{ ft}\)

Exercise 3 – Solve a right triangle

\(c=\sqrt{49+576}=25\). \(\arctan\frac{7}{24}\approx 16.26^{\circ}\), and the other angle is \(90^{\circ}-16.26^{\circ}=73.74^{\circ}\).

\(25\); \(16.26^{\circ}\) and \(73.74^{\circ}\)

Exercise 4 – Law of Sines (AAS)

Solved triangle.

\(C=180^{\circ}-40^{\circ}-60^{\circ}=80^{\circ}\). \(b=\frac{10\sin60^{\circ}}{\sin40^{\circ}}\approx 13.47\). \(c=\frac{10\sin80^{\circ}}{\sin40^{\circ}}\approx 15.32\).

\(C=80^{\circ}\), \(b\approx 13.47\), \(c\approx 15.32\)

Exercise 5 – Law of Sines (ASA-like)

\(B=60^{\circ}\). \(a=\frac{12\sin50^{\circ}}{\sin70^{\circ}}\approx 9.78\). \(b=\frac{12\sin60^{\circ}}{\sin70^{\circ}}\approx 11.06\).

\(B=60^{\circ}\), \(a\approx 9.78\), \(b\approx 11.06\)

Exercise 6 – SSA with one triangle

\(\sin B=\frac{10\sin40^{\circ}}{15}\approx 0.4285\), so \(B\approx 25.37^{\circ}\). The other value \(180^{\circ}-25.37^{\circ}\) is too large since \(A+B>180^{\circ}\). \(C\approx 114.63^{\circ}\), \(c=\frac{15\sin C}{\sin40^{\circ}}\approx 21.21\).

\(B\approx 25.37^{\circ}\), \(C\approx 114.63^{\circ}\), \(c\approx 21.21\)

Exercise 7 – Ambiguous case: two triangles

B1 and B2.

\(\sin B=\frac{10\sin35^{\circ}}{7}\approx 0.8194\). So \(B\approx 55.0^{\circ}\) or \(180^{\circ}-55.0^{\circ}=125.0^{\circ}\). Both leave a positive third angle (\(35+125.0< 180\)).

Two triangles: \(B\approx 55.0^{\circ}\) or \(125.0^{\circ}\)

Exercise 8 – No triangle

\(\sin B=\frac{10\sin50^{\circ}}{5}\approx 1.532>1\).

No triangle exists

Exercise 9 – Law of Cosines (SAS)

\(c^{2}=64+121-2(8)(11)\cos48^{\circ}\approx 67.23\), so \(c\approx 8.20\).

\(c\approx 8.20\)

Exercise 10 – Law of Cosines (SSS)

The largest angle is opposite \(12\): \(\cos C=\frac{49+81-144}{2\cdot7\cdot9}=-\frac{14}{126}\approx-0.1111\), so \(C\approx 96.38^{\circ}\).

\(96.38^{\circ}\)

Exercises 11 and up

Exercise 11 – All three angles

Angle opposite 5: \(\cos A=\frac{36+49-25}{84}\), \(A\approx 44.42^{\circ}\). Angle opposite 6: \(\cos B=\frac{25+49-36}{70}\), \(B\approx 57.12^{\circ}\). \(C=180^{\circ}-44.42^{\circ}-57.12^{\circ}=78.46^{\circ}\).

\(44.42^{\circ}\), \(57.12^{\circ}\), \(78.46^{\circ}\)

Exercise 12 – Area from two sides

\(\frac12(9)(14)\sin35^{\circ}\approx 36.14\).

\(36.14\) square units

Exercise 13 – Heron's formula

\(s=\frac{13+14+15}{2}=21\). Area \(=\sqrt{21\cdot8\cdot7\cdot6}=\sqrt{7056}\).

84 square units

Exercise 14 – Choose the law

(a) SAS: Law of Cosines. (b) Two angles and a side: Law of Sines. (c) SSS: Law of Cosines. (d) SSA with a pair \(a\), \(A\): Law of Sines.

(a) Cosines; (b) Sines; (c) Cosines; (d) Sines

Exercise 15 – Width of a pond

AB is opposite the 62 degree angle.

\(AB^{2}=120^{2}+95^{2}-2(120)(95)\cos62^{\circ}\approx 12721\), so \(AB\approx 112.8\).

\(112.8\text{ ft}\)

Exercise 16 – Two ships

The angle at port is 115 – 40 = 75 degrees.

The angle at the port is \(115^{\circ}-40^{\circ}=75^{\circ}\). \(d^{2}=144+81-2(12)(9)\cos75^{\circ}\approx 169.1\), so \(d\approx 13.0\).

\(13.0\text{ mi}\)

Exercise 17 – Height of a tree

Distance from the base: h/tan 30 and h/tan 45.

The distances to the base are \(\frac{h}{\tan30^{\circ}}=h\sqrt3\) and \(\frac{h}{\tan45^{\circ}}=h\). Their difference is \(40\): \(h(\sqrt3-1)=40\), so \(h=\frac{40}{\sqrt3-1}\approx 54.6\).

\(54.6\text{ ft}\)

Exercise 18 – Triangular garden

\(s=36\). Area \(=\sqrt{36\cdot18\cdot12\cdot6}=216\). Since \(18^{2}+24^{2}=900=30^{2}\), the largest angle is a right angle.

\(216\text{ ft}^{2}\); \(90^{\circ}\)

Exercise 19 – Roof truss

\(\cos C=\frac{100+100-144}{200}=0.28\), so \(C\approx 73.74^{\circ}\). The base angles are equal: \(\frac{180^{\circ}-73.74^{\circ}}{2}=53.13^{\circ}\).

Apex \(73.74^{\circ}\); base angles \(53.13^{\circ}\)

Exercise 20 – Area from angles and a side

\(C=70^{\circ}\). \(b=\frac{10\sin70^{\circ}}{\sin40^{\circ}}\approx 14.62\). Area \(=\frac12ab\sin C=\frac12(10)(14.62)\sin70^{\circ}\approx 68.7\).

\(68.7\) square units

Key facts behind this laws sines cosines answers

Key facts

Law of Sines: a/sin A = b/sin B = c/sin C.
Law of Cosines: c^2 = a^2 + b^2 – 2ab cos C.
SSA can give 0, 1 or 2 triangles (the ambiguous case).
Area = (1/2) a b sin C; Heron: sqrt(s(s – a)(s – b)(s – c)) with s = (a + b + c)/2.
The angles of a triangle add to 180 degrees; the largest angle is opposite the longest side.

How to use this laws 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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