Question

The cosine of the obtuse angle formed by medians drawn from the vertices of the acute angles of an isosceles right-angled triangle is $-\frac{k}{5}$, where $k=$

Answer 1

Let $ABC$ be the right-angled triangle, with $AB=BC=a,$ so that $AC=\sqrt{2}a.$

Let the medians $AD$ and $CE$ intersect at $O.$

Then, $CO=AO=\frac{2}{3}AD=\frac{2}{3}\sqrt{a^2+\frac{a^2}{4}}=\frac{a\sqrt{5}}{3}$

$\cos \left(\angle AOC\right)=\frac{\frac{5a^2}{9}+\frac{5a^2}{9}-2a^2}{2\times \frac{\sqrt{5}a}{3}\times \frac{\sqrt{5}a}{3}}=-\frac{4}{5}$

$\therefore k=4$