## 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: 36

#### Answer

The angles of the triangle are as follows: $A = 24.1^{\circ}, B = 30.8^{\circ},$ and $C = 125.1^{\circ}$ The lengths of the sides are as follows: $a = 4.0~ft, b = 5.0~ft,$ and $c = 8.0~ft$

#### Work Step by Step

We can use the law of cosines to find $B$: $b^2 = a^2+c^2-2ac~cos~B$ $2ac~cos~B = a^2+c^2-b^2$ $cos~B = \frac{a^2+c^2-b^2}{2ac}$ $B = arccos(\frac{a^2+c^2-b^2}{2ac})$ $B = arccos(\frac{4.0^2+8.0^2-5.0^2}{(2)(4.0)(8.0)})$ $B = arccos(0.859375)$ $B = 30.8^{\circ}$ We can use the law of cosines to find $C$: $c^2 = a^2+b^2-2ab~cos~C$ $2ab~cos~C = a^2+b^2-c^2$ $cos~C = \frac{a^2+b^2-c^2}{2ab}$ $C = arccos(\frac{a^2+b^2-c^2}{2ab})$ $C = arccos(\frac{4.0^2+5.0^2-8.0^2}{(2)(4.0)(5.0)})$ $C = arccos(-0.575)$ $C = 125.1^{\circ}$ We can find angle $A$: $A+B+C = 180^{\circ}$ $A = 180^{\circ}-B-C$ $A = 180^{\circ}-30.8^{\circ}-125.1^{\circ}$ $A = 24.1^{\circ}$

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