Problem - 2983
Let $ABCD$ be a convex quadrilateral and the two diagonals intersect at point $O$. If $AC = 12, BD=5$ and $\angle{AOB}=2\angle{BOC}$, find the area of $S_{ABCD}$
$$\angle{BOC}=60^\circ\implies S_{ABCD}=\frac{1}{2} \cdot AC \cdot BD \cdot\sin 60^\circ= \boxed{15\sqrt{3}}$$