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 316: 50

Answer

$53^{\circ}$

Work Step by Step

We can use the law of cosines to find the unknown angle. The law of cosines is: $a^{2}=b^{2}+c^{2}-2bc\cos \beta$ where $a,b,c$ are the three sides of the triangle while $\beta$ is the angle opposite the side $a$. Substituting the values in the formula and solving: $a^{2}=b^{2}+c^{2}-2bc\cos \beta$ $16^{2}=13^{2}+20^{2}-2(13)(20)\cos \beta$ $256=169+400-520\cos \beta$ $256=569-520\cos \beta$ $256-569=-520\cos \beta$ $-313=-520\cos \beta$ $-520\cos \beta=-313$ $\cos \beta=\frac{-313}{-520}$ $\cos \beta=\frac{313}{520}$ $\beta=\cos^{-1} \frac{313}{520}$ $\beta=52.99^{\circ}\approx53^{\circ}$
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