Finding the third side of a triangle given the area
Solution 1:
By cosine rule,
\begin{align}
c^2&=a^2+b^2-2ab\cos\gamma
\tag{1}\label{1}
,
\end{align}
And from the formula for the area
\begin{align}
2S&=ab\sin\gamma
,\\
\sin\gamma&=\frac {2S}{ab}
,
\end{align}
so, with \begin{align} \cos\gamma&=\pm\sqrt{1-\sin^2\gamma} = \pm\sqrt{1-\frac {4S^2}{a^2b^2}} =\pm\frac{\sqrt{a^2b^2-4S^2}}{ab} \end{align}
\eqref{1} becomes
\begin{align} c^2&=a^2+b^2\pm2\sqrt{a^2b^2-4S^2} . \end{align}