Answer
$B=30{}^\circ,a\approx 316.0$ and $b\approx 174.3$.
Work Step by Step
First we will find the value of C
Properties of a triangle:
The sum of three angles is $A+B+C=180{}^\circ $
$\begin{align}
& A+B+C=180{}^\circ \\
& 115{}^\circ +B+35{}^\circ =180{}^\circ \\
& B=180{}^\circ -150{}^\circ \\
& B=30{}^\circ
\end{align}$
Now, we will find the remaining sides using the ratio
$\frac{c}{\sin C}$,or $\frac{200}{\sin 35{}^\circ }$,
Now, we will use the law of sines to find a.
$\begin{align}
& \frac{a}{\sin A}=\frac{c}{\sin C} \\
& \frac{a}{\sin 115{}^\circ }=\frac{200}{\sin 35{}^\circ } \\
& a=\frac{200\sin 115{}^\circ }{\sin 35{}^\circ } \\
& a\approx 316.0
\end{align}$
Using the law of sines again, we will find b.
$\begin{align}
& \frac{b}{\sin B}=\frac{c}{\sin C} \\
& \frac{b}{\sin 30{}^\circ }=\frac{200}{\sin 35{}^\circ } \\
& b=\frac{200\sin 30{}^\circ }{\sin 35{}^\circ } \\
& b\approx 174.3
\end{align}$
