## 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: 15

#### Answer

The angles of the triangle are as follows: $A = 73.8^{\circ}, B = 53.1^{\circ},$ and $C = 53.1^{\circ}$ The sides of the triangle are as follows: $a = 12, b = 10,$ and $c = 10$

#### Work Step by Step

Let $a = 12,$ let $b = 10,$ and let $c=10$ We can use the law of cosines to find $B$: $b^2 = a^2+c^2-2ac~cos~B$ $2ac~cos~B = a^2+c^2-b^2$ $cos~B = \frac{a^2+c^2-b^2}{2ac}$ $B = arccos(\frac{a^2+c^2-b^2}{2ac})$ $B = arccos(\frac{12^2+10^2-10^2}{(2)(12)(10)})$ $B = arccos(0.6)$ $B = 53.1^{\circ}$ By symmetry, angle $C = 53.1^{\circ}$ We can find angle $A$: $A+B+C = 180^{\circ}$ $A = 180^{\circ}-B-C$ $A = 180^{\circ}-53.1^{\circ}-53.1^{\circ}$ $A = 73.8^{\circ}$

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