In the right triangle $ABC$ with shorter side $AC$ the hypotenuse $AB$ has length $12$. Denote $T$ its centroid and $D$ the feet of altitude from the vertex $C$. Determine the size of its inner angle at the vertex $B$ for which the triangle $DTC$ has the greatest possible area.
A fairly neat cartbash:
Let $C$ be the origin, and lines $CA$ and $CB$ be the $y$ and $x-$ axes respectively. Let the vertex angle at $B$ be $\theta$. We immediately get the coordinates-
$$T = (4\cos{\theta},4\sin{\theta})$$$$D = (6\sin{2\theta}\sin{\theta},6\sin{2\theta}\cos{\theta})$$Giving the area of $DTC$ after some calculations as $6|\sin{4\theta}|$. This is maximum when $\theta$ is $\boxed{\frac{\pi}{8}}$