Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.3 The Law of Cosines - 7.3 Exercises: 17

Answer

The angles of the triangle are as follows: $A = 56.7^{\circ}, B = 55^{\circ},$ and $C = 68.3^{\circ}$ The lengths of the sides are as follows: $a = 90, b = 88.2,$ and $c = 100$

Work Step by Step

Let $a=90$, and let $c = 100$. We can use the law of cosines to find $b$: $b^2 = a^2+c^2-2ac~cos~B$ $b = \sqrt{a^2+c^2-2ac~cos~B}$ $b = \sqrt{90^2+100^2-(2)(90)(100)~cos~55^{\circ}}$ $b = \sqrt{7775.62}$ $b = 88.2$ We can use the law of sines to find $A$: $\frac{b}{sin~B} = \frac{a}{sin~A}$ $sin~A = \frac{a~sinB}{b}$ $A = arcsin(\frac{a~sinB}{b})$ $A = arcsin(\frac{90~sin(55^{\circ})}{88.2})$ $A = arcsin(0.83587)$ $A = 56.7^{\circ}$ We can find angle $C$: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-56.7^{\circ}-55^{\circ}$ $C = 68.3^{\circ}$
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