## Precalculus (6th Edition) Blitzer

The missing values are: $C=55{}^\circ,b\approx 10.4\text{ and }c\approx 10.5$
For any triangle, The law of sines states: $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\operatorname{sinC}}$ Using the angle sum property, we will obtain: \begin{align} & A+B+C=180{}^\circ \\ & 70{}^\circ +55{}^\circ +C=180{}^\circ \\ & C=180{}^\circ -125{}^\circ \\ & C=55{}^\circ \end{align} Using the law of sines we get, \begin{align} & \frac{a}{\sin A}=\frac{c}{\operatorname{sinC}} \\ & \frac{12}{\sin 70{}^\circ }=\frac{c}{\sin 55{}^\circ } \\ & c\cdot \sin 70{}^\circ =12\cdot \sin 55{}^\circ \\ & c=\frac{12\cdot \sin 55{}^\circ }{\sin 70{}^\circ } \end{align} This gives $c\approx 10.5$ to the nearest tenth. Also, \begin{align} & \frac{a}{\sin A}=\frac{b}{\sin B} \\ & \frac{12}{\sin 70{}^\circ }=\frac{b}{\sin 55{}^\circ } \\ & b\cdot \sin 70{}^\circ =12\cdot \sin 55{}^\circ \\ & b=\frac{12\cdot \sin 55{}^\circ }{\sin 70{}^\circ } \end{align} This gives $b\approx 10.5$ to the nearest tenth. So, $C=55{}^\circ,b\approx 10.4\text{ and }c\approx 10.5$. Hence, $C=55{}^\circ,b\approx 10.4\text{ and }c\approx 10.5$.