Precalculus (6th Edition) Blitzer

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

Chapter 6 - Section 6.1 - The Law of Sines - Exercise Set - Page 720: 31


There is no triangle possible with the provided measurements.

Work Step by Step

Use the ratio $\frac{a}{\sin A}$ or $\frac{9.3}{\sin 18{}^\circ }$. Now, use the law of sines to find B. $\begin{align} & \frac{b}{\sin B}=\frac{a}{\sin A} \\ & \frac{9.3}{\sin 18{}^\circ }=\frac{41}{\sin B} \\ & \sin B=\frac{41\sin 18{}^\circ }{9.3} \\ & B\approx 1.36 \end{align}$ As the sine ratio can never exceed 1, there is no angle B for which $\sin B=1.36$. There is no triangle with the provided measurements.
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