## Precalculus (6th Edition) Blitzer

The missing values are $B=9{}^\circ,C=148{}^\circ \text{ and }c\approx 73.6$.
For any triangle, The law of sines states that: $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\operatorname{sinC}}$ The law of cosines states that: \begin{align} & {{a}^{2}}={{b}^{2}}+{{c}^{2}}-2\cdot b\cdot c\cdot \cos A \\ & {{b}^{2}}={{a}^{2}}+{{c}^{2}}-2\cdot a\cdot c\cdot \cos B \\ & {{c}^{2}}={{a}^{2}}+{{b}^{2}}-2\cdot a\cdot b\cdot \cos C \\ \end{align} Now from the law of sines we get: \begin{align} & \frac{a}{\sin A}=\frac{b}{\sin B} \\ & \frac{54.3}{\sin 23{}^\circ }=\frac{22.1}{\sin B} \\ & 54.3\left( \sin B \right)=22.1\left( \sin 23{}^\circ \right) \\ & \sin B=\frac{22.1\left( \sin 23{}^\circ \right)}{54.3} \end{align} We will simplify it further to obtain the measure of angle B. \begin{align} & \sin B=0.15 \\ & B={{\sin }^{-1}}\left( 0.15 \right) \\ & B\approx 9{}^\circ \end{align} Now using the angle sum property: \begin{align} & A+B+C=180{}^\circ \\ & 23{}^\circ +9{}^\circ +C=180{}^\circ \\ & C=180{}^\circ -32{}^\circ \\ & C=148{}^\circ \end{align} Finally, from the law of sines: \begin{align} & \frac{a}{\sin A}=\frac{c}{\sin C} \\ & \frac{54.3}{\sin 23{}^\circ }=\frac{c}{\sin 148{}^\circ } \\ & c\cdot \sin 23{}^\circ =54.3\left( \sin 148{}^\circ \right) \\ & c=\frac{54.3\left( \sin 148{}^\circ \right)}{\sin 23{}^\circ } \end{align} This gives $c\approx 73.6$. So, $B=9{}^\circ,C=148{}^\circ \text{ and }c\approx 73.6\text{ }$.