Answer
The two sets of possible values are $B=55{}^\circ,\text{ }C=86{}^\circ,\text{ and }c\approx 31.7\text{ or }B=125{}^\circ,\text{ }C=16{}^\circ,\text{ and }c\approx 8.8$.
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 law of sines we get,
$\begin{align}
& \frac{a}{\sin A}=\frac{b}{\sin B} \\
& \frac{20}{\sin 39{}^\circ }=\frac{26}{\sin B} \\
& 20\cdot \sin B=26\cdot \sin 39{}^\circ \\
& \sin B=\frac{26\cdot \sin 39{}^\circ }{20}
\end{align}$
We will simplify it further to get the value of B.
$\begin{align}
& \sin B=0.81 \\
& B={{\sin }^{-1}}\left( 0.81 \right) \\
& B\approx 55{}^\circ,125{}^\circ
\end{align}$
Using the angle sum property for both the possible values of angle B, we get:
For $B=55{}^\circ $,
$\begin{align}
& 39{}^\circ +55{}^\circ +C=180{}^\circ \\
& C=180{}^\circ -94{}^\circ \\
& C=86{}^\circ
\end{align}$
For $B=125{}^\circ $,
$\begin{align}
& 39{}^\circ +125{}^\circ +C=180{}^\circ \\
& C=180{}^\circ -164{}^\circ \\
& C=16{}^\circ
\end{align}$
Now, obtain the value of side c by using both the possible values of angle C in the law of sines.
For $C=16{}^\circ $,
$\begin{align}
& \frac{20}{\sin 39{}^\circ }=\frac{c}{\sin 16{}^\circ } \\
& c\cdot \sin 39{}^\circ =20\cdot \sin 16{}^\circ \\
& c=\frac{20\cdot \sin 16{}^\circ }{\sin 39{}^\circ } \\
& c\approx 8.8
\end{align}$
For $C=86{}^\circ $,
$\begin{align}
& \frac{20}{\sin 39{}^\circ }=\frac{c}{\sin 86{}^\circ } \\
& c\cdot \sin 39{}^\circ =20\cdot \sin 86{}^\circ \\
& c=\frac{20\cdot \sin 86{}^\circ }{\sin 39{}^\circ } \\
& c\approx 31.7
\end{align}$
So, the two possible sets of combinations are as follows:
$\begin{align}
& B=55{}^\circ,\text{ }C=86{}^\circ,\text{ and }c\approx 31.7\text{ } \\
& B=125{}^\circ,\text{ }C=16{}^\circ,\text{ and }c\approx 8.8
\end{align}$
