## Precalculus (6th Edition) Blitzer

The missing values are $A=72{}^\circ,C=42{}^\circ \text{ and }b\approx 16.3$.
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: \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} The law of cosines gives: \begin{align} & {{b}^{2}}={{a}^{2}}+{{c}^{2}}-2\cdot a\cdot c\cdot \cos B \\ & ={{17}^{2}}+{{12}^{2}}-2\cdot 17\cdot 12\cdot \cos 66{}^\circ \\ & =289+144-165.9 \\ & =267.1 \end{align} This gives $b\approx 16.3$ to the nearest tenth. Using the law of sines we get, \begin{align} & \frac{a}{\sin A}=\frac{c}{\sin C} \\ & \frac{17}{\sin 107{}^\circ }=\frac{12}{\sin C} \\ & 17\cdot \sin C=12\cdot \sin 107{}^\circ \\ & \sin C=\frac{12\cdot \sin 107{}^\circ }{17} \end{align} We will simplify it further \begin{align} & \sin C=0.67 \\ & C={{\sin }^{-1}}\left( 0.67 \right) \\ & C\approx 42{}^\circ \end{align} Using the angle sum property we will get, \begin{align} & A+B+C=180{}^\circ \\ & A+66{}^\circ +42{}^\circ =180{}^\circ \\ & A=180{}^\circ -108{}^\circ \\ & A=72{}^\circ \end{align} So, $A=72{}^\circ,C=42{}^\circ \text{ and }b\approx 16.3$. Hence, $A=72{}^\circ,C=42{}^\circ \text{ and }b\approx 16.3$.