Answer
The missing value are: $A=39{}^\circ,B=78{}^\circ \text{ and }C=63{}^\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:
$\begin{align}
& {{a}^{2}}={{b}^{2}}+{{c}^{2}}-2\cdot b\cdot c\cdot \cos A \\
& {{b}^{2}}={{a}^{2}}+{{c}^{2}}-2\cdot a\cdot c\cdot \cos B \\
& {{c}^{2}}={{a}^{2}}+{{b}^{2}}-2\cdot a\cdot b\cdot \cos C \\
\end{align}$
Use the law of cosines to obtain:
$\begin{align}
& {{b}^{2}}={{a}^{2}}+{{c}^{2}}-2\cdot a\cdot c\cdot \cos B \\
& {{40.2}^{2}}={{26.1}^{2}}+{{36.5}^{2}}-2\cdot \left( 26.1 \right)\cdot \left( 36.5 \right)\cdot \cos B \\
& \cos B=\frac{\left( {{26.1}^{2}}+{{36.5}^{2}} \right)-{{40.2}^{2}}}{2\cdot \left( 26.1 \right)\cdot \left( 36.5 \right)} \\
& \cos B\approx 0.20
\end{align}$
We will simplify it further to get the measure of angle B.
$\begin{align}
& \cos B\approx 0.20 \\
& B\approx {{\cos }^{-1}}\left( 0.20 \right) \\
& B\approx 78{}^\circ
\end{align}$
Using the law of sines we get,
$\begin{align}
& \frac{a}{\sin A}=\frac{b}{\sin B} \\
& \frac{26.1}{\sin A}=\frac{40.2}{\sin 78{}^\circ } \\
& 40.2\left( \sin A \right)=26.1\left( \sin 78{}^\circ \right) \\
& \sin A=\frac{26.1\left( \sin 78{}^\circ \right)}{40.2}
\end{align}$
We will simplify it further to obtain the measure of angle A.
$\begin{align}
& \sin A=0.63 \\
& A={{\sin }^{-1}}\left( 0.63 \right) \\
& A\approx 39{}^\circ
\end{align}$
Using the angle sum property we will obtain the measure of angle C as:
$\begin{align}
& A+B+C=180{}^\circ \\
& 39{}^\circ +78{}^\circ +C=180{}^\circ \\
& C=180{}^\circ -117{}^\circ \\
& C=63{}^\circ
\end{align}$
So,
$A=39{}^\circ,B=78{}^\circ \text{ and }C=63{}^\circ $.
Hence, $A=39{}^\circ,B=78{}^\circ \text{ and }C=63{}^\circ $.
