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 313: 12

Answer

$\theta = 30^{\circ}$

Work Step by Step

Let $a = \sqrt{3}$, let $b=1$, and let $c = 1$. We can use the law of cosines to find $\theta$: $c^2 = a^2+b^2-2ab~cos~\theta$ $2ab~cos~\theta = a^2+b^2-c^2$ $cos~\theta = \frac{a^2+b^2-c^2}{2ab}$ $\theta = arccos(\frac{a^2+b^2-c^2}{2ab})$ $\theta = arccos(\frac{(\sqrt{3})^2+1^2-1^2}{(2)(\sqrt{3})(1)})$ $\theta = arccos(\frac{3}{2\sqrt{3}})$ $\theta = arccos(\frac{\sqrt{3}}{2})$ $\theta = 30^{\circ}$
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