## Precalculus (6th Edition) Blitzer

The missing values are: $B=25{}^\circ,C=115{}^\circ \text{ and c}\approx 8.5$.
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 sates 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} Using the law of sines we get, \begin{align} & \frac{a}{\sin A}=\frac{b}{\sin B} \\ & \frac{6}{\sin 40{}^\circ }=\frac{4}{\sin B} \\ & 6\cdot \sin B=4\cdot \sin 40{}^\circ \\ & \sin B=\frac{4\cdot \sin 40{}^\circ }{6} \end{align} We will simplify it further to get the measure of angle B. \begin{align} & \sin B=0.42 \\ & B={{\sin }^{-1}}\left( 0.42 \right) \\ & B\approx 25{}^\circ \\ \end{align} Using the angle sum property we will obtain the measure of angle C. \begin{align} & A+B+C=180{}^\circ \\ & 40{}^\circ +25{}^\circ +C=180{}^\circ \\ & C=180{}^\circ -65{}^\circ \\ & C=115{}^\circ \end{align} Finally use the law if sines: \begin{align} & \frac{a}{\sin A}=\frac{c}{\sin C} \\ & \frac{6}{\sin 40{}^\circ }=\frac{c}{\sin 115{}^\circ } \\ & 6\cdot \sin 115{}^\circ =c\cdot \sin 40{}^\circ \\ & c=\frac{6\cdot \sin 115{}^\circ }{\sin 40{}^\circ } \end{align} This gives $c\approx 8.5$. So, $B=25{}^\circ,C=115{}^\circ \text{ and c}\approx 8.5$. Hence, $B=25{}^\circ,C=115{}^\circ \text{ and c}\approx 8.5$.