## 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: 23

#### Answer

The angles of the triangle are as follows: $A = 81.8^{\circ}, B = 37.4^{\circ},$ and $C = 60.8^{\circ}$ The lengths of the sides are as follows: $a = 9.3~cm, b = 5.7~cm,$ and $c = 8.2~cm$

#### Work Step by Step

We can use the law of cosines to find $B$: $b^2 = a^2+c^2-2ac~cos~B$ $2ac~cos~B = a^2+c^2-b^2$ $cos~B = \frac{a^2+c^2-b^2}{2ac}$ $B = arccos(\frac{a^2+c^2-b^2}{2ac})$ $B = arccos(\frac{9.3^2+8.2^2-5.7^2}{(2)(9.3)(8.2)})$ $B = arccos(0.7949)$ $B = 37.4^{\circ}$ We can use the law of cosines to find $C$: $c^2 = a^2+b^2-2ab~cos~C$ $2ab~cos~C = a^2+b^2-c^2$ $cos~C = \frac{a^2+b^2-c^2}{2ab}$ $C = arccos(\frac{a^2+b^2-c^2}{2ab})$ $C = arccos(\frac{9.3^2+5.7^2-8.2^2}{(2)(9.3)(5.7)})$ $C = arccos(0.488)$ $C = 60.8^{\circ}$ We can find angle $A$: $A+B+C = 180^{\circ}$ $A = 180^{\circ}-B-C$ $A = 180^{\circ}-37.4^{\circ}-60.8^{\circ}$ $A = 81.8^{\circ}$

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