Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.3 The Law of Cosines - 7.3 Exercises - Page 318: 62

Answer

$\theta = 14.25^{\circ}$

Work Step by Step

Let $a$ be the length of the line from the origin to the point (8,6). $a = \sqrt{8^2+6^2} = 10$ Let $b$ be the length of the line from the origin to the point (12,5). $b = \sqrt{12^2+5^2} = 13$ Let $c$ be the length of the line from the point (8,6) to the point (12,5). $c = \sqrt{(12-8)^2+(5-6)^2} = \sqrt{17}$ We can use the law of cosines to find $\theta$: $c^2 = a^2+b^2-2ab~cos~\theta$ $2ab~cos~\theta = a^2+b^2-c^2$ $cos~\theta = \frac{a^2+b^2-c^2}{2ab}$ $\theta = arccos(\frac{a^2+b^2-c^2}{2ab})$ $\theta = arccos(\frac{10^2+13^2-\sqrt{17}^2}{(2)(10)(13)})$ $\theta = arccos(0.96923)$ $\theta = 14.25^{\circ}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.