Answer
The values $A=55{}^\circ,B=27{}^\circ \text{ and }C=98{}^\circ \text{ or }A=55{}^\circ,B=45{}^\circ \text{ and }C=80{}^\circ $.
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;
$\begin{align}
& {{a}^{2}}={{b}^{2}}+{{c}^{2}}-2\cdot b\cdot c\cdot \cos A \\
& {{b}^{2}}={{a}^{2}}+{{c}^{2}}-2\cdot a\cdot c\cdot \cos B \\
& {{c}^{2}}={{a}^{2}}+{{b}^{2}}-2\cdot a\cdot b\cdot \cos C \\
\end{align}$
From the law of cosines:
$\begin{align}
& {{a}^{2}}={{b}^{2}}+{{c}^{2}}-2\cdot b\cdot c\cdot \cos A \\
& {{117}^{2}}={{66}^{2}}+{{142}^{2}}-2\cdot 66\cdot 142\cdot \cos A \\
& \cos A=\frac{\left( {{66}^{2}}+{{142}^{2}} \right)-{{117}^{2}}}{2\cdot 66\cdot 142} \\
& \cos A=\frac{24520-13689}{18744}
\end{align}$
We will simplify it further
$\begin{align}
& \cos A=0.57 \\
& A={{\cos }^{-1}}\left( 0.57 \right) \\
& A\approx 55{}^\circ
\end{align}$
Using the law of sines we get,
$\begin{align}
& \frac{a}{\sin A}=\frac{c}{\sin C} \\
& \frac{117}{\sin 55{}^\circ }=\frac{142}{\sin C} \\
& 117\cdot \sin C=142\cdot \sin 55{}^\circ \\
& \sin C=\frac{142\cdot \sin 55{}^\circ }{117}
\end{align}$
Proceed further to obtain the measure of angle C
$\begin{align}
& \sin C=0.99 \\
& C={{\sin }^{-1}}\left( 0.99 \right) \\
& C\approx 98{}^\circ,80{}^\circ
\end{align}$
Using the angle sum property for both values of angle C.
For $C=98{}^\circ $,
$\begin{align}
& 55{}^\circ +B+98{}^\circ =180{}^\circ \\
& B=180{}^\circ -153{}^\circ \\
& B=27{}^\circ
\end{align}$
For $C=80{}^\circ $,
$\begin{align}
& 55{}^\circ +B+80{}^\circ =180{}^\circ \\
& B=180{}^\circ -135{}^\circ \\
& B=45{}^\circ
\end{align}$
So, the two combinations for the triangles are:
$\begin{align}
& A=55{}^\circ,B=27{}^\circ \text{ and }C=98{}^\circ \\
& A=55{}^\circ,B=45{}^\circ \text{ and }C=80{}^\circ \\
\end{align}$
Hence, $A=55{}^\circ,B=27{}^\circ \text{ and }C=98{}^\circ \text{ or }A=55{}^\circ,B=45{}^\circ \text{ and }C=80{}^\circ $.
