How to find an angle of a non-right angle triangle when given two sides and an area? How would I go about finding an angles of a non-right angled triangle when given the area and two of its sides.

For example:

In the triangle $ABC$, $a = 5$, $b = 6$ and the area is $11~\text{cm}^2$. Find the size of Angle $C$.

Thanks for all of your help! :-)

Note:

What is wrong with the current method I am using? (method below)

\begin{align*}
Area &amp; = \frac{ab\sin C}{2}\\
\Rightarrow 2Area &amp; = ab\sin C\\
\Rightarrow \frac{2Area}{ab} &amp; = \sin C\\
\sin C &amp; = \frac{11^2 \cdot 2}{5 \cdot 6}\\
\Rightarrow \sin C &amp; = 8.06
\end{align*}
You still need to do this ~ C=sin-1(11/15) 
Because you are trying to find C alone. 
Correct me if I'm wrong.
You still need to do this ~ C=sin-1(11/15) 
Because you are trying to find C alone. 
Correct me if I'm wrong.