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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:

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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 $\alpha=\angle A$

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  • $\begingroup$ $\angle COB= 2\alpha$, but $\angle CBO=\angle OCB= 90 - \alpha$ $\endgroup$ – asdf Dec 20 '17 at 9:29
  • $\begingroup$ @asdf: Thank you. I edited. $\endgroup$ – Roman83 Dec 20 '17 at 9:42
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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. $$\frac{\pi}{2}-\frac{\arccos\frac{19}{20}+\arccos\frac{1}{4}}{2}.$$

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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 $\gamma$.

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