## Precalculus (6th Edition) Blitzer

The missing values are $B=3{}^\circ,C=15{}^\circ \text{ and }a\approx 59$.
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} Use the law of cosines to obtain: \begin{align} & {{a}^{2}}={{b}^{2}}+{{c}^{2}}-2\cdot b\cdot c\cdot \cos A \\ & {{a}^{2}}={{11.2}^{2}}+{{48.2}^{2}}-2\left( 11.2 \right)\left( 48.2 \right)\cos 162{}^\circ \\ & {{a}^{2}}\approx 347.5 \\ & a\approx \sqrt{347.5} \end{align} This gives $a\approx 59$. Using the law of sines we get, \begin{align} & \frac{a}{\sin A}=\frac{b}{\sin B} \\ & \frac{59}{\sin 162{}^\circ }=\frac{11.2}{\sin B} \\ & 59\cdot \sin B=11.2\left( \sin 162{}^\circ \right) \\ & \sin B=\frac{11.2\left( \sin 162{}^\circ \right)}{59} \end{align} We will simplify it further to obtain the value of angle B as follows: \begin{align} & \sin B=0.05 \\ & B={{\sin }^{-1}}\left( 0.05 \right) \\ & B\approx 3{}^\circ \\ \end{align} Using the angle sum property we will obtain the measure of angle C. \begin{align} & A+B+C=180{}^\circ \\ & 162{}^\circ +3{}^\circ +C=180{}^\circ \\ & C=180{}^\circ -165{}^\circ \\ & C=15{}^\circ \end{align} So, $B=3{}^\circ,C=15{}^\circ \text{ and }a\approx 59.0$. Hence, $B=3{}^\circ,C=15{}^\circ \text{ and }a\approx 59$.