In the triangle $ABC$ $R = \frac56 BH = \frac52OH$. Find the angles $ACB$ or $BAC$ 
In the triangle $ABC$, the height $BH$ is drawn, the point $O$ is the center of the circle circumscribed about it, the length of its radius $R$. Find the smallest of the angles $ACB$ and $BAC$, expressed in radians, if it is known that $R = \frac56 BH = \frac52OH$

My work so far:

1) In triangle $BOH$ $BO=R, BH=\frac65R, OH=\frac25R$. Then I can to find $\angle BOH, \angle BHO$ and $\angle OBH$
2) I proved that $\angle ABH= \angle OBC=90^{\circ}-\alpha$, where $\alpha=\angle A$

Another approach.  Let $S$ be the foot of $O$ on $AC$.  We know that $AC=2AS$. First calculate $\angle BHO$. Then from $\triangle HSO$ find $HS$ and $OS$. From $\triangle OSC$ we now can find $SC$ (Pythagoras) which is equal to $AS$. Now $AH=SC-SH$ and find $\angle\alpha$ from $\triangle AHB$.
As we know $AC$, use $2R=\dfrac{AC}{\sin(\beta)}$ to find $\beta$.  As we know $\alpha$ and $\beta$, we also know $\gamma$.
