Answer
The required values are $B=19{}^\circ,b\approx 9.6$, and $c\approx 23.1$.
Work Step by Step
At first we will find B.
Property of a triangle:
Sum of three angles is $A+B+C=180{}^\circ $
$\begin{align}
& A+B+C=180{}^\circ \\
& 33{}^\circ +B+128{}^\circ =180{}^\circ \\
& B=180{}^\circ -161{}^\circ \\
& B=19{}^\circ
\end{align}$
Now, to find the remaining sides, we will use the ratio $\frac{a}{\sin A}$ or $\frac{16}{\sin 33{}^\circ }$,
Now, we will use the law of sines to find b.
$\begin{align}
& \frac{b}{\sin B}=\frac{a}{\sin A} \\
& \frac{b}{\sin 19{}^\circ }=\frac{16}{\sin 33{}^\circ } \\
& b=\frac{16\sin 19{}^\circ }{\sin 33{}^\circ } \\
& b=9.6
\end{align}$
Again use the law of sines to find c
$\begin{align}
& \frac{c}{\sin C}=\frac{a}{\sin A} \\
& \frac{c}{\sin 128{}^\circ }=\frac{16}{\sin 33{}^\circ } \\
& c=\frac{16\sin 128{}^\circ }{\sin 33{}^\circ } \\
& c=23.1
\end{align}$
The solution is $B=19{}^\circ,b\approx 9.6$, and $c\approx 23.1$.
