## Precalculus (6th Edition) Blitzer

The missing quantities for the triangle are $c=39.5,B={{54}^{{}^\circ }},C={{92}^{{}^\circ }}$ or $c=13.7,B={{126}^{{}^\circ }},C={{20}^{{}^\circ }}$.
By using the sine rule, $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$ For B, \begin{align} & \sin B=\frac{b\times \sin A}{a} \\ & \sin B=\frac{32\times \sin {{34}^{{}^\circ }}}{22} \\ & \sin B=0.8134 \\ & B\approx {{54}^{{}^\circ }},{{126}^{{}^\circ }} \end{align} Now, use the angle sum property to find the value of C. \begin{align} & A+B+C={{180}^{{}^\circ }} \\ & {{34}^{{}^\circ }}+{{54}^{{}^\circ }}+C={{180}^{{}^\circ }} \\ & C={{92}^{{}^\circ }} \end{align} Or \begin{align} & {{34}^{{}^\circ }}+{{126}^{{}^\circ }}+C={{180}^{{}^\circ }} \\ & C={{180}^{{}^\circ }}-{{160}^{{}^\circ }} \\ & C={{20}^{{}^\circ }} \end{align} Applying sine rule for $C={{92}^{{}^\circ }}$ \begin{align} & c=22\frac{\sin {{92}^{{}^\circ }}}{\sin {{34}^{{}^\circ }}} \\ & =39.5 \end{align} Then, apply sine rule for $C={{20}^{{}^\circ }}$ \begin{align} & c=22\frac{\sin {{20}^{{}^\circ }}}{\sin {{34}^{{}^\circ }}} \\ & =13.7 \end{align} Hence, the missing quantities for the triangle are $c=39.5,B={{54}^{{}^\circ }},C={{92}^{{}^\circ }}$ or $c=13.7,B={{126}^{{}^\circ }},C={{20}^{{}^\circ }}$.