## Precalculus (6th Edition) Blitzer

There is one triangle and the solution is ${{B}_{1}}\left( \text{or }B \right)\approx 29{}^\circ,\ C\approx 111{}^\circ,\ \text{ and }\ c\approx 29.0$.
We will use the law of sines to find B: \begin{align} & \frac{a}{\sin A}=\frac{b}{\sin B} \\ & \frac{20}{\sin 40{}^\circ }=\frac{15}{\sin B} \\ & \sin B=\frac{15\sin 40{}^\circ }{20} \\ & \approx 0.4821 \end{align} There are two angles between 0° and 180° for which $\sin B=0.4821$ is possible: ${{B}_{1}}\approx 29{}^\circ$ As we know, the sine function is positive in the second quadrant, so the second angle is \begin{align} & {{B}_{2}}\approx 180{}^\circ -29{}^\circ \\ & \approx 151{}^\circ \end{align} Here, ${{B}_{2}}$ is impossible, as $40{}^\circ +151{}^\circ =191{}^\circ$. Now, find $C$ using ${{B}_{1}}$ and $A=40{}^\circ$. By the angle sum property, \begin{align} & C=180{}^\circ -{{B}_{1}}-A \\ & \approx 180{}^\circ -29{}^\circ -40{}^\circ \\ & \approx 111{}^\circ \end{align} Using the law of sines we will find side c: \begin{align} & \frac{c}{\sin C}=\frac{a}{\sin A} \\ & \frac{c}{\sin 111{}^\circ }=\frac{20}{\sin 40{}^\circ } \\ & c=\frac{20\sin 111{}^\circ }{\sin 40{}^\circ } \\ & \approx 29.0 \end{align}