Answer
The missing values are $A=72{}^\circ,C=42{}^\circ \text{ and }b\approx 16.3$.
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:
$\begin{align}
& {{a}^{2}}={{b}^{2}}+{{c}^{2}}-2\cdot b\cdot c\cdot \cos A \\
& {{b}^{2}}={{a}^{2}}+{{c}^{2}}-2\cdot a\cdot c\cdot \cos B \\
& {{c}^{2}}={{a}^{2}}+{{b}^{2}}-2\cdot a\cdot b\cdot \cos C \\
\end{align}$
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, $A=72{}^\circ,C=42{}^\circ \text{ and }b\approx 16.3$.
