I have been asked to finish this one - I am your humble servant (only this time lol) so I shall be pleased to do so.
$$\begin{array}{rlll}
\frac{12}{sinB}&=&\frac{11}{sin30}\\\\
\frac{sinB}{12}&=&\frac{sin30}{11}\\\\
SinB&=&\frac{12sin30}{11}\\\\
B&=&asin\left( {\frac{12sin30}{11}}\right)\\\\
B&=&33.056^0\qquad &\mbox{First quadrant solution correct to 3 dec places} \\\\
B&=&180-33.056=146.944\qquad &\mbox{Second quadrant solution correct to 3 dec places}\\\\
\end{array}$$
$$If B is $33^0$ then \\
$C=180-30-33=117^0$\\\\
If B is $147^0$ then \\
$C=180-30-147=3^0$\\\\
So the 2 possible solutions are $
I've have constructed the triangle to show you that there are exactly 2 solutions solutions.
This triangle is drawn to scale.
We need to find angle B first.
You can do this using the sine rule but remember, since sin is positive in the first and second quadrant there will be an accute angle B and an obtuse angle B - Two possible solutions for B
Once you have angle B you can find Angle C by using angle sum of a triangle. Once agian there will be 2 solutions.
You try finding this these angles.
If you get stuck again put another post after this one to ask for more help.
I have been asked to finish this one - I am your humble servant (only this time lol) so I shall be pleased to do so.
$$\begin{array}{rlll}
\frac{12}{sinB}&=&\frac{11}{sin30}\\\\
\frac{sinB}{12}&=&\frac{sin30}{11}\\\\
SinB&=&\frac{12sin30}{11}\\\\
B&=&asin\left( {\frac{12sin30}{11}}\right)\\\\
B&=&33.056^0\qquad &\mbox{First quadrant solution correct to 3 dec places} \\\\
B&=&180-33.056=146.944\qquad &\mbox{Second quadrant solution correct to 3 dec places}\\\\
\end{array}$$
$$If B is $33^0$ then \\
$C=180-30-33=117^0$\\\\
If B is $147^0$ then \\
$C=180-30-147=3^0$\\\\
So the 2 possible solutions are $