## Precalculus (6th Edition) Blitzer

There is one triangle and the solution is ${{C}_{1}}\left( \text{or }C \right)\approx 37{}^\circ,\ B\approx 7{}^\circ,\ \text{ and }\ b\approx 10.1$.
We will use the law of sines to find C: \begin{align} & \frac{a}{\sin A}=\frac{c}{\sin C} \\ & \frac{57.5}{\sin 136{}^\circ }=\frac{49.8}{\sin C} \\ & \sin C=\frac{49.8\sin 136{}^\circ }{57.5} \\ & \approx 0.6016 \end{align} There are two angles between 0° and 180° for which $\sin C=0.6016$ is possible: ${{C}_{1}}\approx 37{}^\circ$ Since we know that the sine function is positive in the second quadrant, the second angle is \begin{align} & {{C}_{2}}\approx 180{}^\circ -37{}^\circ \\ & \approx 143{}^\circ \end{align} Here, ${{C}_{2}}$ is impossible, as $136{}^\circ +143{}^\circ =279{}^\circ$. Now, find $B$ using ${{C}_{1}}$ and $A=136{}^\circ$. Using the angle sum property, we get \begin{align} & B=180{}^\circ -{{C}_{1}}-A \\ & \approx 180{}^\circ -37{}^\circ -136{}^\circ \\ & \approx 7{}^\circ \end{align} Using the law of sines, we will find side b: \begin{align} & \frac{b}{\sin B}=\frac{a}{\sin A} \\ & \frac{b}{\sin 7{}^\circ }=\frac{57.5}{\sin 136{}^\circ } \\ & b=\frac{57.5\sin 7{}^\circ }{\sin 136{}^\circ } \\ & \approx 10.1 \end{align}