## Precalculus (6th Edition) Blitzer

The missing values are: $C=120{}^\circ,a\approx 45\text{ and }b\approx 33.2$.
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} Using the angle sum property : \begin{align} & A+B+C=180{}^\circ \\ & 35{}^\circ +25{}^\circ +C=180{}^\circ \\ & C=180{}^\circ -60{}^\circ \\ & C=120{}^\circ \end{align} Using the law of sines we get, \begin{align} & \frac{a}{\sin A}=\frac{c}{\operatorname{sinC}} \\ & \frac{a}{\sin 35{}^\circ }=\frac{68}{\sin 120{}^\circ } \\ & a\cdot \sin 120{}^\circ =68\cdot \sin 35{}^\circ \\ & a=\frac{68\cdot \sin 35{}^\circ }{\sin 120{}^\circ } \end{align} This gives $a\approx 45.0$ to the nearest tenth. Also, \begin{align} & \frac{a}{\sin A}=\frac{b}{\sin B} \\ & \frac{45}{\sin 35{}^\circ }=\frac{b}{\sin 25{}^\circ } \\ & b\cdot \sin 35{}^\circ =45\cdot \sin 25{}^\circ \\ & b=\frac{45\cdot \sin 25{}^\circ }{\sin 35{}^\circ } \end{align} This gives $b\approx 33.2$ to the nearest tenth. So, $C=120{}^\circ,a\approx 45\text{ and }b\approx 33.2$. Hence, $C=120{}^\circ,a\approx 45\text{ and }b\approx 33.2$.