How to find an angle of a non-right angle triangle when given two sides and an area?

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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 & = \frac{ab\sin C}{2}\\ \Rightarrow 2Area & = ab\sin C\\ \Rightarrow \frac{2Area}{ab} & = \sin C\\ \sin C & = \frac{11^2 \cdot 2}{5 \cdot 6}\\ \Rightarrow \sin C & = 8.06 \end{align*}

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1 Answer

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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.

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