## 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: 20

#### Answer

The angles of the triangle are as follows: $A = 44.9^{\circ}, B = 106.8^{\circ},$ and $C = 28.3^{\circ}$ The lengths of the sides are as follows: $a = 4.21~in, b = 5.71~in,$ and $c = 2.83~in$

#### Work Step by Step

We can use the law of cosines to find $c$: $c^2 = a^2+b^2-2ab~cos~C$ $c = \sqrt{a^2+b^2-2ab~cos~C}$ $c = \sqrt{(4.21~in)^2+(5.71~in)^2-(2)(4.21~in)(5.71~in)~cos~28.3^{\circ}}$ $c = \sqrt{7.996~in^2}$ $c = 2.83~in$ We can use the law of cosines to find $B$: $b^2 = a^2+c^2-2ac~cos~B$ $2ac~cos~B = a^2+c^2-b^2$ $cos~B = \frac{a^2+c^2-b^2}{2ac}$ $B = arccos(\frac{a^2+c^2-b^2}{2ac})$ $B = arccos(\frac{(4.21)^2+(2.83)^2-(5.71)^2}{(2)(4.21)(2.83)})$ $B = arccos(-0.288)$ $B = 106.8^{\circ}$ We can find angle $A$: $A+B+C = 180^{\circ}$ $A = 180^{\circ}-B-C$ $A = 180^{\circ}-106.8^{\circ}-28.3^{\circ}$ $A = 44.9^{\circ}$

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