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 $\alpha=\angle A$
 A: The hint.
In $\Delta HOB$ we know that $OH=\frac{2}{5}R$, $BH=\frac{6}{5}R$ and $BO=R$.
Thus, by law of cosines we obtain $$\cos\cos\measuredangle HBO HBO=\frac{1+\frac{36}{25}-\frac{4}{25}}{2\cdot\frac{6}{5}},$$
which gives
$$\cos\measuredangle HBO=\frac{19}{20}.$$
In another hand, $\cos\measuredangle HBO=|\alpha-\gamma|$, which gives
$$\cos(\alpha-\gamma)=\frac{19}{20}.$$
Now, $$BH=c\sin\alpha=2R\sin\alpha\sin\gamma.$$
Thus, $$R=\frac{5}{6}\cdot2R\sin\alpha\sin\gamma$$ or
$$\sin\alpha\sin\gamma=\frac{3}{5}.$$
I hope the rest is smooth because
$$\cos(\alpha+\gamma)=\frac{19}{20}-2\cdot\frac{3}{5}=-\frac{1}{4}.$$
I got the following value.
$$\frac{\pi}{2}-\frac{\arccos\frac{19}{20}+\arccos\frac{1}{4}}{2}.$$
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 $\gamma$.
