Geometry: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281159
ISBN 13: 978-0-13328-115-6

Chapter 8 - Right Triangles and Trigonometry - 8-6 Law of Cosines - Practice and Problem-Solving Exercises - Page 530: 7


$QS = 10.1$

Work Step by Step

First, we want to know what angle is opposite the side in question. The angle that is opposite to the side we are looking for is $\angle R$, so let's plug in what we know into the formula for the law of cosines: $(QS)^2 = 11^2 + 16^2 - 2(11)(16)$ cos $38^{\circ}$ Evaluate exponents first, according to order of operations: $(QS)^2 = 121 + 256 - 2(11)(16)$ cos $38^{\circ}$ Add to simplify on the right side of the equation: $(QS)^2 = 377 - 2(11)(16)$ cos $38^{\circ}$ Evaluate the right side of the equation: $(QS)^2 = 99.6$ Take the square root of both sides to solve: $QS = 10.1$
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