## Precalculus (6th Edition) Blitzer

The missing values are: $A=43{}^\circ,a\approx 171.9\text{ and }b\approx 241.0$.
For any triangle, The law of sines states: $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\operatorname{sinC}}$ Then use the angle sum property and obtain: \begin{align} & A+B+C=180{}^\circ \\ & A+107{}^\circ +30{}^\circ =180{}^\circ \\ & A=180{}^\circ -137{}^\circ \\ & A=43{}^\circ \end{align} Using the law of sines we get, \begin{align} & \frac{a}{\sin A}=\frac{c}{\operatorname{sinC}} \\ & \frac{a}{\sin 43{}^\circ }=\frac{126}{\sin 30{}^\circ } \\ & a\cdot \sin 30{}^\circ =126\cdot \sin 43{}^\circ \\ & a=\frac{126\cdot \sin 43{}^\circ }{\sin 30{}^\circ } \end{align} This gives $a\approx 171.9$ to the nearest tenth. Also, \begin{align} & \frac{b}{\sin B}=\frac{c}{\sin C} \\ & \frac{b}{\sin 107{}^\circ }=\frac{126}{\sin 30{}^\circ } \\ & b\cdot \sin 30{}^\circ =126\cdot \sin 107{}^\circ \\ & b=\frac{126\cdot \sin 107{}^\circ }{\sin 30{}^\circ } \end{align} This gives $b\approx 241.0$ to the nearest tenth. So, $A=43{}^\circ,a\approx 171.9\text{ and }b\approx 241.0$.