Answer
The missing values are: $C=120{}^\circ,a\approx 45\text{ and }b\approx 33.2$.
Work Step by Step
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, $C=120{}^\circ,a\approx 45\text{ and }b\approx 33.2$.
