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: 20


There is one triangle and the solution is ${{C}_{1}}\left( \text{or }C \right)\approx 37{}^\circ,\ B\approx 7{}^\circ,\ \text{ and }\ b\approx 10.1$.

Work Step by Step

We will use the law of sines to find C: $\begin{align} & \frac{a}{\sin A}=\frac{c}{\sin C} \\ & \frac{57.5}{\sin 136{}^\circ }=\frac{49.8}{\sin C} \\ & \sin C=\frac{49.8\sin 136{}^\circ }{57.5} \\ & \approx 0.6016 \end{align}$ There are two angles between 0° and 180° for which $\sin C=0.6016$ is possible: ${{C}_{1}}\approx 37{}^\circ $ Since we know that the sine function is positive in the second quadrant, the second angle is $\begin{align} & {{C}_{2}}\approx 180{}^\circ -37{}^\circ \\ & \approx 143{}^\circ \end{align}$ Here, ${{C}_{2}}$ is impossible, as $136{}^\circ +143{}^\circ =279{}^\circ $. Now, find $B$ using ${{C}_{1}}$ and $A=136{}^\circ $. Using the angle sum property, we get $\begin{align} & B=180{}^\circ -{{C}_{1}}-A \\ & \approx 180{}^\circ -37{}^\circ -136{}^\circ \\ & \approx 7{}^\circ \end{align}$ Using the law of sines, we will find side b: $\begin{align} & \frac{b}{\sin B}=\frac{a}{\sin A} \\ & \frac{b}{\sin 7{}^\circ }=\frac{57.5}{\sin 136{}^\circ } \\ & b=\frac{57.5\sin 7{}^\circ }{\sin 136{}^\circ } \\ & \approx 10.1 \end{align}$
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