Find angles in a right triangle if $\tan\theta = \frac34$, where $\theta$ is the angle between medians Find angles in right triangle if it is known that $\tan \theta = {3\over4}$, where $\theta$ is the angle between catheti medians. I have tried drawing orthogonal projections and making new right triangles with that same angle but it did not seem to lead anywhere, as well as trying to find something that could be useful for sine theorem, but again got nothing good.
Have tried to make new right triangles from centroid, but did not get anything useful.
Any help would be much appreciated. As it was a problem on a competition, calculators were not allowed



The length of one median is $\sqrt{a^2 + 4b^2}$ and the other is  $\sqrt{4a^2 + b^2}$
The intersection of the medians split each other with a ratio of $2:1$
Law of sines
$\frac {\sin x}{a} = \frac {\sin y}{\frac 13 \sqrt{a^2 + 4b^2}}$
$\sin x = \frac 35$
$\sin y = \frac {\sqrt{a^2 + 4b^2}}{5a}$
Aslo $\sin y = \frac b{\sqrt{4a^2 + b^2}}$
$\frac {\sqrt{a^2 + 4b^2}}{5a} = \frac b{\sqrt{4a^2 + b^2}}\\
5ab = \sqrt {4a^4 + 17a^2b^2 + 4b^4}\\
25a^2b^2 =  4a^4 + 17a^2b^2 + 4b^4\\
4a^4 - 8a^2b^2 + b^4) = 0\\
4(a^2 - b^2)^2 = 0$
$a = b$ 
you have an isoceles right triangle.
