## Precalculus (6th Edition) Blitzer

There are two triangles and the solutions are ${{B}_{1}}\approx 27{}^\circ,\ {{C}_{1}}\approx 133{}^\circ,\ \text{ and }\ {{c}_{1}}\approx 64.2$, and ${{B}_{2}}\approx 153{}^\circ,\ {{C}_{2}}\approx 7{}^\circ,\ \text{ and }\ {{c}_{2}}\approx 10.7$.
We will use the law of sines to find B: \begin{align} & \frac{a}{\sin A}=\frac{b}{\sin B} \\ & \frac{30}{\sin 20{}^\circ }=\frac{40}{\sin B} \\ & \sin B=\frac{40\sin 20{}^\circ }{30} \\ & \approx 0.4560 \end{align} There are two angles between 0° and 180° for which $\sin B\approx 0.4560$ is possible: ${{B}_{1}}\approx 27{}^\circ$ As we know that the sine function is positive in the second quadrant, the second angle is \begin{align} & {{B}_{2}}\approx 180{}^\circ -27{}^\circ \\ & \approx 153{}^\circ \end{align} So, there are two triangles. For ${{B}_{1}}\approx 27{}^\circ$, \begin{align} & {{C}_{1}}=180{}^\circ -{{B}_{1}}-A \\ & =180{}^\circ -27{}^\circ -20{}^\circ \\ & =133{}^\circ \end{align} For ${{c}_{1}}$, \begin{align} & \frac{{{c}_{1}}}{\sin {{C}_{1}}}=\frac{a}{\sin A} \\ & \frac{{{c}_{1}}}{\sin 133{}^\circ }=\frac{30}{\sin 20{}^\circ } \\ & {{c}_{1}}=\frac{30\sin 133{}^\circ }{\sin 20{}^\circ } \\ & \approx 64.2 \end{align} For ${{B}_{2}}\approx 153{}^\circ$, \begin{align} & {{C}_{2}}=180{}^\circ -{{B}_{2}}-A \\ & =180{}^\circ -153{}^\circ -20{}^\circ \\ & =7{}^\circ \end{align} For ${{c}_{2}}$, \begin{align} & \frac{{{c}_{2}}}{\sin {{C}_{2}}}=\frac{a}{\sin A} \\ & \frac{{{c}_{2}}}{\sin 7{}^\circ }=\frac{30}{\sin 20{}^\circ } \\ & {{c}_{2}}=\frac{30\sin 7{}^\circ }{\sin 20{}^\circ } \\ & \approx 10.7 \end{align} In case of the first triangle, the solution is ${{B}_{1}}\approx 27{}^\circ,\ {{C}_{1}}\approx 133{}^\circ,\ \text{ and }\ {{c}_{1}}\approx 64.2$ In the other triangle, the solution is ${{B}_{2}}\approx 153{}^\circ,\ {{C}_{2}}\approx 7{}^\circ,\ \text{ and }\ {{c}_{2}}\approx 10.7$