The acute angle between the medians drawn through the acute angles of an isosceles right angled triangle is

$\cos^{-1}(\frac{4}{5})$

Let ΔAOB be an isosceles right triangle and let be the acute angle between its medians AD and BE as shown in Fig. Taking O as the origin, let the position vectors of A and B be $\vec a$ and $\vec b$ respectively.

Since $\vec{OA}⊥\vec{OB}$ and $OA = OB$

$∴\vec a.\vec b=0$ and $|\vec a|=|\vec b|$

Now,

$\vec{AD}=\frac{\vec b}{2}-\vec a$ and $\vec{BE}=\frac{\vec a}{2}-\vec b$

$⇒|\vec{AD}|=\left|\frac{\vec b}{2}-\vec a\right|^2$ and $|\vec{BE}|^2\left|\frac{\vec a}{2}-\vec b\right|^2$

$⇒|\vec{AD}|=\frac{1}{4}|\vec b-2\vec a|^2$ and $|BE|^2=\frac{1}{4}|\vec a-2\vec b|^2$

$⇒|\vec{AD}|^2=\frac{1}{4}\left\{|\vec b|^2+4|\vec a|^2-4(\vec a.\vec b)\right\}$

and, $|BE|^2=\frac{1}{4}\left\{|\vec a|^2+4|\vec b|^2-4(\vec a.\vec b)\right\}$

$⇒|\vec{AD}|^2=\frac{5}{4}|\vec a|^2$ and $|BE|^2=\frac{5}{4}|\vec a|^2$  $[∵|\vec a|=|\vec b|\, and\, \vec a.\vec b=0]$

$⇒|\vec{AD}|=\sqrt{\frac{5}{4}}|\vec a|$ and $|BE|=\sqrt{\frac{5}{4}}|\vec a|$

Now,

$\cos θ=\left|\frac{\vec{AD}.\vec{BE}}{|\vec{AD}||\vec{BE}|}\right|$

$⇒\cos θ=\left|\frac{\left(\frac{\vec b}{2}-\vec a\right).\left(\frac{\vec a}{2}-\vec b\right)}{\sqrt{\frac{5}{4}}|\vec a|\sqrt{\frac{5}{4}}|\vec a|}\right|$

$⇒\cos θ=\frac{1}{5|\vec a|^2}|(\vec b-2\vec a)-(\vec a-2\vec b)|$

$⇒\cos θ=\frac{1}{5|\vec a|^2}|-2|\vec a|^22|\vec b|^2+5(\vec a.\vec b)|$

$⇒\cos θ=\frac{4|\vec a|^2}{5|\vec a|^2}$   $[∵|\vec a|=|\vec b|\,and\,\vec a.\vec b=0]$

$⇒θ=\cos^{-1}(\frac{4}{5})$