In triangle ABC, medians AD and BE are drawn. If $\mathrm{AD}=4,\mathrm{\angle }\mathrm{DAB}=\frac{\mathrm{\pi }}{6}$ and , $\mathrm{\angle }\mathrm{ABE}=\frac{\mathrm{\pi }}{3}$ then the area of the $△\mathrm{ABC}$ is

Draw the figure with the given data

We know that

$\begin{array}{rcl}Area of ∆ABC& =& 3\left(Area of ∆ABG\right)\\ & =& 3\left(\frac{1}{2}AG·BG\right)\\ & =& \frac{3}{2}\left(\frac{2}{3}AD\right)\left(AG\tan 30°\right)\\ & =& \frac{3}{2}\left(\left(\frac{8}{3}\right)·\frac{8}{3\sqrt{3}}\right)\\ & =& \frac{32}{3\sqrt{3}}\end{array}$