Area of a triangle

Let $s = 6$ and $\theta$ be the included angle between these two equal sides (and opposite the third side). This angle value is computed via:

$A = \frac{1}{2}s^2 \sin \theta \Rightarrow \theta = \arcsin(\frac{2A}{s^2}) = \arcsin(\frac{2 \cdot 9\sqrt{3}}{6^2}) = \arcsin(\frac{\sqrt{3}}{2}) = \frac{\pi}{3}$ .

The triangle is therefore equilateral, and the third side has length of 6.