Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 6 - Section 6.2 - The Law of Cosines - Exercise Set - Page 731: 46

Answer

a. $23.1\ mi$. b. $S\ 68^\circ \ E$
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Work Step by Step

a. Using the figure as shown, in triangle ABC, we have the angle $B=90^\circ-45^\circ=45^\circ$ Using the Law of Cosines, we have $b^2=30^2+12^2-2(30)(12)cos(45^\circ)\approx534.9$ which gives $b\approx23.1\ mi$. b. Using the Law of Sines, we have $\frac{sinA}{12}=\frac{sin45^\circ}{23.1}$ which gives $sinA=\frac{12sin45^\circ}{23.1}\approx0.3673$ Thus, $A=asin(0.3673)\approx22^\circ$ and the bearing from A to C is $S\ (90^\circ-22^\circ)\ E$ or $S\ 68^\circ \ E$
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