Precalculus (6th Edition) Blitzer

There are two triangles and the solutions are ${{B}_{1}}\approx 54{}^\circ,\ {{C}_{1}}\approx 89{}^\circ,\ \text{ and }\ {{c}_{1}}\approx 19.9$, and ${{B}_{2}}\approx 126{}^\circ,\ {{C}_{2}}\approx 17{}^\circ,\ \text{ and }\ {{c}_{2}}\approx 5.8$.
We will use the law of sines to find B: \begin{align} & \frac{a}{\sin A}=\frac{b}{\sin B} \\ & \frac{12}{\sin 37{}^\circ }=\frac{16.1}{\sin B} \\ & \sin B=\frac{16.1\sin 37{}^\circ }{12} \\ & \sin B\approx 0.8074 \end{align} There are two angles between 0° and 180° for which $\sin B\approx 0.8074$ is possible: ${{B}_{1}}=54{}^\circ$ Since the sine function is positive in the second quadrant, the second angle is \begin{align} & {{B}_{2}}=180{}^\circ -54{}^\circ \\ & =126{}^\circ \end{align} So, there are two triangles. In case of the first triangle, \begin{align} & {{C}_{1}}=180{}^\circ -{{B}_{1}}-A \\ & =180{}^\circ -54{}^\circ -37{}^\circ \\ & =89{}^\circ \end{align} In case of the second triangle, \begin{align} & {{C}_{2}}=180{}^\circ -{{B}_{2}}-A \\ & =180{}^\circ -126{}^\circ -37{}^\circ \\ & =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 89{}^\circ }=\frac{12}{\sin 37{}^\circ } \\ & {{c}_{1}}=\frac{12\sin 89{}^\circ }{\sin 37{}^\circ } \\ & {{c}_{1}}\approx 19.9 \end{align} For ${{c}_{2}}$, \begin{align} & \frac{{{c}_{2}}}{\sin {{C}_{2}}}=\frac{a}{\sin A} \\ & \frac{{{c}_{2}}}{\sin 17{}^\circ }=\frac{12}{\sin 37{}^\circ } \\ & {{c}_{2}}=\frac{12\sin 17{}^\circ }{\sin 37{}^\circ } \\ & {{c}_{2}}\approx 5.8 \end{align} In first triangle, the solution is ${{B}_{1}}\approx 54{}^\circ,\ {{C}_{1}}\approx 89{}^\circ,\ \text{ and }\ {{c}_{1}}\approx 19.9$ In the other triangle, the solution is ${{B}_{2}}\approx 126{}^\circ,\ {{C}_{2}}\approx 17{}^\circ,\ \text{ and }\ {{c}_{2}}\approx 5.8$