Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.2 The Ambiguous Case of the Law of Sines - 7.2 Exercises - Page 304: 11

Answer

$B = 45^{\circ}$

Work Step by Step

We can use the law of sines to find the angle $B$: $\frac{a}{sin~A} = \frac{b}{sin~B}$ $sin~B = \frac{b~sin~A}{a}$ $sin~B = \frac{(2)~sin~(60^{\circ})}{\sqrt{6}}$ $sin~B = \frac{(2)~(\frac{\sqrt{3}}{2})}{\sqrt{6}}$ $sin~B = \frac{\sqrt{2}}{2}$ $B = arcsin(\frac{\sqrt{2}}{2})$ $B = 45^{\circ}$
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