## Trigonometry (10th Edition)

Published by Pearson

# Chapter 7 - Applications of Trigonometry and Vectors - Section 7.3 The Law of Cosines - 7.3 Exercises - Page 314: 31

#### Answer

The angles of the triangle are as follows: $A = 64.6^{\circ}, B = 74.8^{\circ},$ and $C = 40.6^{\circ}$ The lengths of the sides are as follows: $a = 8.92~in, b = 9.53~in,$ and $c = 6.43~in$

#### Work Step by Step

We can use the law of cosines to find $b$: $b^2 = a^2+c^2-2ac~cos~B$ $b = \sqrt{a^2+c^2-2ac~cos~B}$ $b = \sqrt{(8.92~in)^2+(6.43~in)^2-(2)(8.92~in)(6.43~in)~cos~74.8^{\circ}}$ $b = \sqrt{90.835~in^2}$ $b = 9.53~in$ We can use the law of cosines to find $C$: $c^2 = a^2+b^2-2ab~cos~C$ $2ab~cos~C = a^2+b^2-c^2$ $cos~C = \frac{a^2+b^2-c^2}{2ab}$ $C = arccos(\frac{a^2+b^2-c^2}{2ab})$ $C = arccos(\frac{8.92^2+9.53^2-6.43^2}{(2)(8.92)(9.53)})$ $C = arccos(0.759)$ $C = 40.6^{\circ}$ We can find angle $A$: $A+B+C = 180^{\circ}$ $A = 180^{\circ}-B-C$ $A = 180^{\circ}-74.8^{\circ}-40.6^{\circ}$ $A = 64.6^{\circ}$

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