## 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: 28

#### Answer

The angles of the triangle are as follows: $A = 50.1^{\circ}, B = 85.1^{\circ},$ and $C = 44.8^{\circ}$ The lengths of the sides are as follows: $a = 324~m, b = 421~m,$ and $c = 298~m$

#### Work Step by Step

We can use the law of cosines to find $B$: $b^2 = a^2+c^2-2ac~cos~B$ $2ac~cos~B = a^2+c^2-b^2$ $cos~B = \frac{a^2+c^2-b^2}{2ac}$ $B = arccos(\frac{a^2+c^2-b^2}{2ac})$ $B = arccos(\frac{324^2+298^2-421^2}{(2)(324)(298)})$ $B = arccos(0.0856)$ $B = 85.1^{\circ}$ We can use the law of cosines to find $C$: $c^2 = a^2+b^2-2ab~cos~C$ $2ab~cos~C = a^2+b^2-c^2$ $cos~C = \frac{a^2+b^2-c^2}{2ab}$ $C = arccos(\frac{a^2+b^2-c^2}{2ab})$ $C = arccos(\frac{324^2+421^2-298^2}{(2)(324)(421)})$ $C = arccos(0.709)$ $C = 44.8^{\circ}$ We can find angle $A$: $A+B+C = 180^{\circ}$ $A = 180^{\circ}-B-C$ $A = 180^{\circ}-85.1^{\circ}-44.8^{\circ}$ $A = 50.1^{\circ}$

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