Find the acute angles of this right triangle.

Question

I am having trouble finding the acute angles of this triangle. O is the intersection of the medians of the triangle and $OG = \frac{1}{2}OH$. Any suggestions?

Answer 1

Being $O$ the centroid of the right $\triangle ABC$ we have that $\triangle ABO$, $\triangle BCO$ and $\triangle CAO$ have the same area, then \begin{align*} \mbox{Area }BCO&=\mbox{Area }ABO\\ \frac{1}{2}|BC|\times |OH|&=\frac{1}{2}|AB|\times|OG|\\ \implies\frac{|BC|}{|AB|}&=\frac{|OG|}{|OH|}\\ \cos\measuredangle ABC&=\frac{1}{2} \end{align*} Therefore $\measuredangle ABC=\frac{\pi}{3}$ and $\measuredangle CAB=\frac{\pi}{6}$.

Answer 2

You have to note the points H and G are in the circle of diameter OB. Then calculate sin and cos of angle CBO and ABO. With this you have angle CBA and BAC is the complementaire.