Answer
The required values are $B=40{}^\circ,b\approx 20.9,$ and $c\approx 31.8$.
Work Step by Step
At first we will find B.
Property of a triangle:
Sum of three angles is $A+B+C=180{}^\circ $
$\begin{align}
& A+B+C=180{}^\circ \\
& 38{}^\circ +B+102{}^\circ =180{}^\circ \\
& B=180{}^\circ -140{}^\circ \\
& B=40{}^\circ
\end{align}$
Now, to find the remaining sides we will:
Use the ratio $\frac{a}{\sin A}$ or $\frac{20}{\sin 38{}^\circ }$,
Now we will use the law of sines to find b.
$\begin{align}
& \frac{b}{\sin B}=\frac{a}{\sin A} \\
& \frac{b}{\sin 40{}^\circ }=\frac{20}{\sin 38{}^\circ } \\
& b=\frac{20\sin 40{}^\circ }{\sin 38{}^\circ } \\
& b=20.9
\end{align}$
Again use the law of sines to find c
$\begin{align}
& \frac{c}{\sin C}=\frac{a}{\sin A} \\
& \frac{c}{\sin 102{}^\circ }=\frac{20}{\sin 38{}^\circ } \\
& c=\frac{20\sin 102{}^\circ }{\sin 38{}^\circ } \\
& c=31.8
\end{align}$
The solution is $B=40{}^\circ,b\approx 20.9,$ and $c\approx 31.8$.
