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: 25


There are two triangles and the solutions are ${{B}_{1}}\approx 77{}^\circ,\ {{C}_{1}}\approx 43{}^\circ,\ \text{ and }\ {{c}_{1}}\approx 12.6$, and ${{B}_{2}}\approx 103{}^\circ,\ {{C}_{2}}\approx 17{}^\circ,\ \text{ and }\ {{c}_{2}}\approx 5.4$.

Work Step by Step

We will use the law of sines to find B: $\begin{align} & \frac{a}{\sin A}=\frac{b}{\sin B} \\ & \frac{16}{\sin 60{}^\circ }=\frac{18}{\sin B} \\ & \sin B=\frac{18\sin 60{}^\circ }{16} \\ & \sin B\approx 0.9743 \end{align}$ There are two angles between 0° and 180° for which $\sin B\approx 0.9743$ is possible: ${{B}_{1}}\approx 77{}^\circ $ As we know that the sine function is positive in the second quadrant, so the second angle is $\begin{align} & {{B}_{2}}\approx 180{}^\circ -77{}^\circ \\ & \approx 103{}^\circ \end{align}$ So, there are two triangles. In the first triangle, using the angle sum property we get $\begin{align} & {{C}_{1}}=180{}^\circ -{{B}_{1}}-A \\ & \approx 180{}^\circ -77{}^\circ -60{}^\circ \\ & \approx 43{}^\circ \end{align}$ In the second triangle, using the angle sum property we get $\begin{align} & {{C}_{2}}=180{}^\circ -{{B}_{2}}-A \\ & \approx 180{}^\circ -103{}^\circ -60{}^\circ \\ & \approx 17{}^\circ \end{align}$ Using the law of sines we will find the sides ${{c}_{1}}$ and ${{c}_{2}}$. For ${{c}_{1}}$, $\begin{align} & \frac{{{c}_{1}}}{\sin {{C}_{1}}}=\frac{a}{\sin A} \\ & \frac{{{c}_{1}}}{\sin 43{}^\circ }=\frac{16}{\sin 60{}^\circ } \\ & {{c}_{1}}=\frac{16\sin 43{}^\circ }{\sin 60{}^\circ } \\ & \approx 12.6 \end{align}$ For ${{c}_{2}}$, $\begin{align} & \frac{{{c}_{2}}}{\sin {{C}_{2}}}=\frac{a}{\sin A} \\ & \frac{{{c}_{2}}}{\sin 17{}^\circ }=\frac{16}{\sin 60{}^\circ } \\ & {{c}_{2}}=\frac{16\sin 17{}^\circ }{\sin 60{}^\circ } \\ & \approx 5.4 \end{align}$ In case of the first triangle, the solution is ${{B}_{1}}\approx 77{}^\circ,\ {{C}_{1}}\approx 43{}^\circ,\ \text{ and }\ {{c}_{1}}\approx 12.6$ In the other triangle, the solution is ${{B}_{2}}\approx 103{}^\circ,\ {{C}_{2}}\approx 17{}^\circ,\ \text{ and }\ {{c}_{2}}\approx 5.4$
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