## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 7 - Applications of Trigonometry and Vectors - Section 7.3 The Law of Cosines - 7.3 Exercises - Page 319: 13

#### Answer

The angles are as follows: $A = 121^{\circ}, B = 38.2^{\circ},$ and $C = 20.8^{\circ}$ The lengths of the sides are as follows: $a = 7, b = 5,$ and $c = 3$

#### Work Step by Step

Let $b=5$, and let $c = 3$. We can use the law of cosines to find $a$: $a^2 = b^2+c^2-2bc~cos~A$ $a = \sqrt{b^2+c^2-2bc~cos~A}$ $a = \sqrt{5^2+3^2-(2)(5)(3)~cos~121^{\circ}}$ $a = \sqrt{49.45}$ $a = 7.0$ We can use the law of cosines to find $B$: $b^2 = a^2+c^2-2ac~cos~B$ $2ac~cos~B = a^2+c^2-b^2$ $cos~B = \frac{a^2+c^2-b^2}{2ac}$ $B = arccos(\frac{a^2+c^2-b^2}{2ac})$ $B = arccos(\frac{7^2+3^2-5^2}{(2)(7)(3)})$ $B = arccos(\frac{11}{14})$ $B = 38.2^{\circ}$ We can find angle $C$: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-121^{\circ}-38.2^{\circ}$ $C = 20.8^{\circ}$

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