Question

In a triangle $ABC,AD$ and $BE$ are medians drawn to $BC$ and $CA$ respectively. Given $AD=4,\angle DAB=\frac{\pi }{6}$ and$\angle ABE=\frac{\pi }{3}.$ If the area of triangle $ABC$ is $\frac{p}{\sqrt{3}q},$ where $p$ and $q$ are co-prime, then the value of $p+q$ is

Answer 1

As, $AD=4$
So, $AG=\frac{2}{3}AD$
$\Rightarrow AG=\frac{2}{3}\left(4\right)=\frac{8}{3}$

Now, in $△AGB$
$\frac{BG}{\sin {30}^{\circ }}=\frac{AG}{\sin {60}^{\circ }}$ (Using sine law)
$\Rightarrow BG=\left(\frac{\sin {30}^{\circ }}{\sin {60}^{\circ }}\right)AG$
$\Rightarrow BG=\left(\frac{1/2}{\sqrt{3}/2}\right)\left(\frac{8}{3}\right)=\frac{8}{3\sqrt{3}}$


Now, area $\left(△ABC\right)=3\times$ area $\left(△GAB\right)$
$=3(\frac{1}{2}\times BG\times AG)$
$=3\times \frac{1}{2}\times \frac{8}{3\sqrt{3}}\times \frac{8}{3}$
$=\frac{32}{3\sqrt{3}}=\frac{p}{\sqrt{3}q}$
So, $p=32$ and $q=3$
$p+q=32+3=35$