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: 44


Navigate $N 46^{\circ} W$

Work Step by Step

Using the law of cosines we can plug in the known information: $b^{2} = a^{2}+c^{2} - 2*a*c*cos\beta$ Since $a = 7, c=6$ and $b = 5$ we have: $25 = 49+36-84*cos\beta$ $25 = 85-84*cos\beta$ $cos\beta = 0.714$ $\beta \approx44^{\circ}$ Subtract $44^{\circ}$ from $90^{\circ}$ to get $46^{\circ}$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.