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 - Chapter Test - Page 537: 20


$m \angle K \approx 91.8^{\circ}$

Work Step by Step

We are given the lengths of three sides of a triangle, and we need to find the measure of one angle. We can use the Law of Cosines in this case. First, we want to know what side is opposite the angle in question. The side that is opposite to the angle we are looking for, $\angle K$, has a measure of $13$, so let's plug in what we know into the formula for the law of cosines: $13^2 = 10^2 + 8^2 - 2(10)(8)$ cos $\angle K$ Evaluate exponents first, according to order of operations: $169 = 100 + 64 - 2(10)(8)$ cos $\angle K$ Add to simplify on the right side of the equation: $169 = 164 - 2(10)(8)$ cos $\angle K$ Multiply to simplify: $169 = 164 - 160$ cos $\angle K$ Subtract $949$ from each side of the equation to move constants to the left side of the equation: $5 = -160$ cos $\angle K$ Divide each side by $-972$: cos $\angle K = \frac{5}{-160}$ Take $cos^{-1}$ to solve for $\angle K$: $m \angle K \approx 91.8^{\circ}$
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