## 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: 25

#### Answer

The angles of the triangle are as follows: $A = 42.0^{\circ}, B = 35.9^{\circ},$ and $C = 102.1^{\circ}$ The lengths of the sides are as follows: $a = 42.9~m, b = 37.6~m,$ and $c = 62.7~m$

#### Work Step by Step

We can use the law of cosines to find $B$: $b^2 = a^2+c^2-2ac~cos~B$ $2ac~cos~B = a^2+c^2-b^2$ $cos~B = \frac{a^2+c^2-b^2}{2ac}$ $B = arccos(\frac{a^2+c^2-b^2}{2ac})$ $B = arccos(\frac{42.9^2+62.7^2-37.6^2}{(2)(42.9)(62.7)})$ $B = arccos(0.81)$ $B = 35.9^{\circ}$ We can use the law of cosines to find $C$: $c^2 = a^2+b^2-2ab~cos~C$ $2ab~cos~C = a^2+b^2-c^2$ $cos~C = \frac{a^2+b^2-c^2}{2ab}$ $C = arccos(\frac{a^2+b^2-c^2}{2ab})$ $C = arccos(\frac{42.9^2+37.6^2-62.7^2}{(2)(42.9)(37.6)})$ $C = arccos(-0.21)$ $C = 102.1^{\circ}$ We can find angle $A$: $A+B+C = 180^{\circ}$ $A = 180^{\circ}-B-C$ $A = 180^{\circ}-35.9^{\circ}-102.1^{\circ}$ $A = 42.0^{\circ}$

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