## Precalculus (6th Edition) Blitzer

There is one triangle and the solution is ${{C}_{1}}\left( \text{or }C \right)\approx 52{}^\circ,\ B\approx 65{}^\circ,\ \text{ and }\ b\approx 10.2$.
We will use the law of sines to find C: \begin{align} & \frac{a}{\sin A}=\frac{c}{\sin C} \\ & \frac{10}{\sin 63{}^\circ }=\frac{8.9}{\sin C} \\ & \sin C=\frac{8.9\sin 63{}^\circ }{10} \\ & =0.7930 \end{align} There are two angles between 0° and 180° for which $\sin C=0.7930$ is possible: ${{C}_{1}}=52{}^\circ$ Since we know that the sine function is positive in the second quadrant, the second angle is \begin{align} & {{C}_{2}}=180{}^\circ -52{}^\circ \\ & =128{}^\circ \end{align} Here, ${{C}_{2}}$ is impossible, as $63{}^\circ +128{}^\circ =191{}^\circ$. Now, we will find $B$ using ${{C}_{1}}$ and $A=63{}^\circ$. Using the angle sum property, we get \begin{align} & B=180{}^\circ -{{C}_{1}}-A \\ & =180{}^\circ -52{}^\circ -63{}^\circ \\ & =65{}^\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 65{}^\circ }=\frac{10}{\sin 63{}^\circ } \\ & b=\frac{10\sin 65{}^\circ }{\sin 63{}^\circ } \\ & \approx 10.2 \end{align}