## 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: 24

#### Answer

The angles of the triangle are as follows: $A = 28.6^{\circ}, B = 53.3^{\circ},$ and $C = 98.1^{\circ}$ The lengths of the sides are as follows: $a = 28~ft, b = 47~ft,$ and $c = 58~ft$

#### Work Step by Step

We can use the law of cosines to find $B$: $b^2 = a^2+c^2-2ac~cos~B$ $2ac~cos~B = a^2+c^2-b^2$ $cos~B = \frac{a^2+c^2-b^2}{2ac}$ $B = arccos(\frac{a^2+c^2-b^2}{2ac})$ $B = arccos(\frac{28^2+58^2-47^2}{(2)(28)(58)})$ $B = arccos(0.597)$ $B = 53.3^{\circ}$ We can use the law of cosines to find $C$: $c^2 = a^2+b^2-2ab~cos~C$ $2ab~cos~C = a^2+b^2-c^2$ $cos~C = \frac{a^2+b^2-c^2}{2ab}$ $C = arccos(\frac{a^2+b^2-c^2}{2ab})$ $C = arccos(\frac{28^2+47^2-58^2}{(2)(28)(47)})$ $C = arccos(-0.14)$ $C = 98.1^{\circ}$ We can find angle $A$: $A+B+C = 180^{\circ}$ $A = 180^{\circ}-B-C$ $A = 180^{\circ}-53.3^{\circ}-98.1^{\circ}$ $A = 28.6^{\circ}$

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