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 730: 39

Answer

$61.7\ mi$.
1584663080

Work Step by Step

Step 1. Draw a diagram as shown in the figure. The first ship traveled from point C, $3(14)=42\ mi$, to point A. The second ship traveled from C, $3(10)=30\ mi$, to point B. The angles are given from the bearings. Step 2. In triangle ABC, we have the angle $C=12^\circ+90^\circ+(90^\circ-75^\circ)=117^\circ$. Step 3. Use the Law of Cosines. Letting $AB=c$, we have $c^2=42^2+30^2-2(42)(30)cos(117^\circ)\approx3808$, which gives $c\approx61.7\ mi$. That is, the two ships are about 61.7 miles apart after 3 hours.
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