## 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 324: 61

#### Answer

$\theta = 16.26^{\circ}$

#### Work Step by Step

Let $a$ be the length of the line from the origin to the point (6,8). $a = \sqrt{6^2+8^2} = 10$ Let $b$ be the length of the line from the origin to the point (4,3). $b = \sqrt{4^2+3^2} = 5$ Let $c$ be the length of the line from the point (4,3) to the point (6,8). $c = \sqrt{(6-4)^2+(8-3)^2} = \sqrt{29}$ We can use the law of cosines to find $\theta$: $c^2 = a^2+b^2-2ab~cos~\theta$ $2ab~cos~\theta = a^2+b^2-c^2$ $cos~\theta = \frac{a^2+b^2-c^2}{2ab}$ $\theta = arccos(\frac{a^2+b^2-c^2}{2ab})$ $\theta = arccos(\frac{10^2+5^2-\sqrt{29}^2}{(2)(5)(10)})$ $\theta = arccos(0.96)$ $\theta = 16.26^{\circ}$

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