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: 43

Answer

$N\ 12^\circ \ E$

Work Step by Step

Using the figure given in the exercise, based on the Law of Cosines, we have $7^2=5^2+6^2-2(5)(6)cos(A)$ which gives $cosA=0.2$ Thus angle $A=acos(0.2)\approx78^\circ$ and the bearing from A to C is $N\ (90^\circ-78^\circ)\ E$ or $N\ 12^\circ \ E$
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