## 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: 16

#### Answer

The angles of the triangle are as follows: $A = 22.3^{\circ}, B = 108.2^{\circ},$ and $C = 49.5^{\circ}$ The lengths of the sides are as follows: $a = 4, b = 10,$ and $c = 8$

#### Work Step by Step

Let $a = 4,$ let $b = 10,$ and let $c=8$ We can use the law of cosines to find $B$: $b^2 = a^2+c^2-2ac~cos~B$ $2ac~cos~B = a^2+c^2-b^2$ $cos~B = \frac{a^2+c^2-b^2}{2ac}$ $B = arccos(\frac{a^2+c^2-b^2}{2ac})$ $B = arccos(\frac{4^2+8^2-10^2}{(2)(4)(8)})$ $B = arccos(-\frac{5}{16})$ $B = 108.2^{\circ}$ We can use the law of cosines to find $C$: $c^2 = a^2+b^2-2ab~cos~C$ $2ab~cos~C = a^2+b^2-c^2$ $cos~C = \frac{a^2+b^2-c^2}{2ab}$ $C = arccos(\frac{a^2+b^2-c^2}{2ab})$ $C = arccos(\frac{4^2+10^2-8^2}{(2)(4)(10)})$ $C = arccos(\frac{13}{20})$ $C = 49.5^{\circ}$ We can find angle $A$: $A+B+C = 180^{\circ}$ $A = 180^{\circ}-B-C$ $A = 180^{\circ}-108.2^{\circ}-49.5^{\circ}$ $A = 22.3^{\circ}$

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