## 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 320: 18

#### Answer

The angles of the triangle are as follows: $A = 33.6, B = 50.7^{\circ},$ and $C = 95.7^{\circ}$ The lengths of the sides are as follows: $a = 5, b = 7,$ and $c = 9$

#### Work Step by Step

Let $a = 5,$ let $b = 7,$ and let $c=9$ We can use the law of cosines to find $B$: $b^2 = a^2+c^2-2ac~cos~B$ $2ac~cos~B = a^2+c^2-b^2$ $cos~B = \frac{a^2+c^2-b^2}{2ac}$ $B = arccos(\frac{a^2+c^2-b^2}{2ac})$ $B = arccos(\frac{5^2+9^2-7^2}{(2)(5)(9)})$ $B = arccos(\frac{19}{30})$ $B = 50.7^{\circ}$ We can use the law of cosines to find $C$: $c^2 = a^2+b^2-2ab~cos~C$ $2ab~cos~C = a^2+b^2-c^2$ $cos~C = \frac{a^2+b^2-c^2}{2ab}$ $C = arccos(\frac{a^2+b^2-c^2}{2ab})$ $C = arccos(\frac{5^2+7^2-9^2}{(2)(5)(7)})$ $C = arccos(-0.1)$ $C = 95.7^{\circ}$ We can find angle $A$: $A+B+C = 180^{\circ}$ $A = 180^{\circ}-B-C$ $A = 180^{\circ}-50.7^{\circ}-95.7^{\circ}$ $A = 33.6^{\circ}$

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