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: 9

Answer

$JL \approx 3.9$

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 $\angle K$, so let's plug in what we know into the formula for the law of cosines: $(JL)^2 = 6.4^2 + 2.6^2 - 2(6.4)(2.6)$ cos $10.5^{\circ}$ Evaluate exponents first, according to order of operations: $(JL)^2 = 40.96 + 6.76 - 2(6.4)(2.6)$ cos $10.5^{\circ}$ Add to simplify on the right side of the equation: $(JL)^2 = 47.72 - 2(6.4)(2.6)$ cos $10.5^{\circ}$ Evaluate the right side of the equation: $(JL)^2 \approx 14.9973$ Take the square root of both sides to solve: $JL \approx 3.9$
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