The area of an isosceles triangle is $S$ and the angle between the medians to the legs, facing the base, is $\alpha$. Find the base of the triangle.
$2\sqrt{\dfrac{S\tan\dfrac{\alpha}{2}}{3}}.$
Let $CH$ be the third median of the triangle. The point $M$ lies on $CH$ because the triangle is isosceles, so in $\triangle ABM,$ $MH$ is a median and an altitude so it's an isosceles triangle. Then $\measuredangle AMH=\measuredangle BMH=\dfrac{\alpha}{2}.$ I have studied only trig functions of an acute angle. Can you give me a hint?
Best Answer
The centroid divides the median into the ratio $2:1$ where the $2$ is the side of the angle it is coming from and $1$ is the side of the centroid where the median meets the side. That makes $CM:MH=2:1 \implies MH= \dfrac13 CH \implies CH = 3 MH$.
In $\triangle AHM$ you have $\dfrac{AH}{MH}=\tan\left( \dfrac{\alpha}2 \right) $, $AB=2AH$ and the base-height formula for $S$.