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: 22


There is one triangle and the solution is ${{B}_{1}}\left( \text{or }B \right)\approx 12{}^\circ,\ C\approx 6{}^\circ,\ \text{ and }\ c\approx 2.1$.

Work Step by Step

We will use the law of sines to find B: $\begin{align} & \frac{a}{\sin A}=\frac{b}{\sin B} \\ & \frac{6.1}{\sin 162{}^\circ }=\frac{4}{\sin B} \\ & \sin B=\frac{4\sin 162{}^\circ }{6.1} \\ & \approx 0.2026 \end{align}$ There are two angles between 0° and 180° for which $\sin B\approx 0.2026$ is possible: ${{B}_{1}}\approx 12{}^\circ $ As we know, the sine function is positive in the second quadrant, so the second angle is $\begin{align} & {{B}_{2}}\approx 180{}^\circ -12{}^\circ \\ & \approx 168{}^\circ \end{align}$ Here, ${{B}_{2}}$ is impossible, as $162{}^\circ +168{}^\circ =330{}^\circ $. Now, find $C$ using ${{B}_{1}}$ and $A=162{}^\circ $. Using the angle sum property of triangles: $\begin{align} & C=180{}^\circ -{{B}_{1}}-A \\ & =180{}^\circ -12{}^\circ -162{}^\circ \\ & =6{}^\circ \end{align}$ Using the law of sines to find side c: $\begin{align} & \frac{c}{\sin C}=\frac{a}{\sin A} \\ & \frac{c}{\sin 6{}^\circ }=\frac{6.1}{\sin 162{}^\circ } \\ & c=\frac{6.1\sin 6{}^\circ }{\sin 162{}^\circ } \\ & \approx 2.1 \end{align}$
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.