I can not find a Method for to solve this geometry problem.I don't even know how to start.In fact, I didn't want to add my nonsensical attempts. I looked for a similar question to this question (solved), but unfortunately, I couldn't find it. That's why I need help. I think I don't have enough mathematical information to solve this problem .
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I agree with @Batominovski that this problem, with its haphazard angle measures, is probably intended as an exercise in the trigonometric form of Ceva's Theorem:
$$\frac{\sin\angle OAC}{\sin \angle OAB} \cdot \frac{\sin\angle OBA}{\sin\angle OBC}\cdot\frac{\sin\angle OCB}{\sin\angle OCA}=1 \tag{$\star$}$$
We're given two of these angles ($\angle OAC$ and $\angle OCA$) explicitly. We're also given the isosceles triangle's vertex angle ($\angle ABC$), from which we may deduce base angles $\angle BAC = \angle BCA$; subtracting appropriately, we may consider $\angle OAB$ and $\angle OCB$ known. Thus, $(\star)$ effectively states $$\sin \angle OBA = k \sin\angle OBC \qquad(\text{say, } \sin\theta = k\sin\phi) \tag{1}$$ for a known value of $k$. But we also have $\angle OBA + \angle OBC = \angle ABC$, another known value, so that $$\sin(\angle OBA + \angle OBC) = \sin\angle ABC \qquad (\sin(\theta+\phi)=\sin\psi)\tag{2}$$ "All we need to do" is solve $(1)$ and $(2)$ for $\sin\phi$ in terms of $k$ and $\psi$. Here's a pretty slick way: simply notice that $$ \sin^2\psi = \sin^2\theta + \sin^2\phi + 2 \sin\theta\sin\phi\cos\psi \tag{3} $$ (see image below) so that, replacing $\sin\theta$ with $k\sin\phi$ and noting that all sines are positive, we readily find
$$\sin\phi = \frac{\sin\psi}{\sqrt{k^2 + 1 + 2 k\cos\psi}}\tag{4}$$
Equation $(4)$ solves the crux of the problem. Substituting-in the specific angle values is just tedium. The reader can follow the discussion under $(\star)$, and/or those shown in other answers, to perform the appropriate calculations, using $$\psi := \angle ABC \qquad k := \frac{\sin\angle OAB}{\sin\angle OAC}\cdot\frac{\sin\angle OCA}{\sin\angle OCB}$$
Here's a trigonograph to demonstrate $(3)$, which amounts to applying the Law of Cosines to the (shaded) $\sin\theta$-$\sin\phi$-$\sin\psi$ triangle.
Let's denote $AC = a\,.\,$ Since the triangle $ABC$ is isosceles
$\angle BAC=\angle ACB=0.5\cdot(180-\angle ABC) \approx6.6670^\circ.\qquad\qquad\qquad\qquad\qquad$ $\angle OCA \approx6.0882^\circ.\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ $\angle OAC\approx4.0548^\circ.\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ $\angle BCO=\angle ACB-\angle OCA\approx 0.5788^\circ.\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ $\angle AOC=180^\circ-\angle OAC-\angle OCA\approx 169.8570^\circ.$
We keep an extra digit for maximum precision.
$BC=\dfrac{a}{2\cos(ACB)}\approx0.5034a\;\;$because triangle $ABC$ is isosceles.$\;$ Applying the law of sines for the triangle $AOC$ we get $\dfrac{OC}{\sin (OAC)}=\dfrac{AC}{\sin (AOC)}\,.\quad$ Hence $\;OC=a\cdot\dfrac{\sin (OAC)}{\sin (AOC)}\approx0.40152a\,.\;$
Using the law of cosines for $\triangle BOC\,$:$\quad BO=\sqrt{OC^2+BC^2-2\cdot OC \cdot BC\cdot \cos(BCO)}\approx$
$\approx 0.10198a\,.\;$ Then we apply the law of sines for triangle $BOC:\quad$ $\dfrac{OC}{\sin (OBC)}=\dfrac{BO}{\sin (BCO)}\,\;$ thus obtaning $\;\sin (OBC)=\dfrac{OC}{BO}\sin (BCO)\approx0.039773\,.\;\,$Finally we arrive at $\angle OBC=\sin^{-1}0.039773\approx2.279^\circ.\;$ ($\;$last digit is not precise, and it can be said
that $\angle OBC\approx2.28^\circ.)$ That's it. The problem is fully solved. $$\quad$$ $\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$* * * * * * * * * * * *$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ $$\quad$$ Now let's retrace all the steps and condense the whole computation into a single formula: $$\sin (OBC)=\frac{OC \cdot \sin (BCO)}{\sqrt{OC^2+BC^2-2\cdot OC \cdot BC\cdot \cos (BCO)}}=$$
$$=\dfrac{a\cdot\dfrac{\sin (OAC)}{\sin (AOC)}\sin (BCO)}{\sqrt{\bigg(a\cdot \dfrac{\sin (OAC)}{\sin (AOC)}\bigg)^2+\bigg(\dfrac{a}{2\cos(ACB)}\bigg)^2-\;2a\cdot \dfrac{\sin (OAC)}{\sin (AOC)}\cdot \dfrac{a}{2\cos(ACB)}\cdot \cos (BCO)}}$$ $\qquad\qquad\qquad\qquad\qquad\qquad$ Now we denote the angles with Greek letters
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\angle ACB=\alpha\qquad(\approx 6.6670^\circ)\qquad\qquad\qquad\qquad$ $\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\angle BCO=\beta\qquad(\approx 0.5788^\circ)\qquad\qquad\qquad\qquad$ $\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\angle AOC=\gamma\qquad(\approx 169.8570^\circ)\qquad\qquad\qquad\qquad$ $\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\angle OAC=\phi\qquad(\approx 4.0548^\circ)\qquad\qquad\qquad\qquad\qquad$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$and our expression becomes neat:
$$\sin (OBC)=\dfrac{a\cdot\dfrac{\sin \phi}{\sin \gamma}\sin \beta}{\sqrt{\bigg(a\cdot \dfrac{\sin \phi}{\sin \gamma}\bigg)^2+\bigg(\dfrac{a}{2\cos\alpha}\bigg)^2-\;2a\cdot \dfrac{\sin \phi}{\sin \gamma}\cdot \dfrac{a}{2\cos\alpha}\cdot \cos \beta}}$$ This expression can be simplified by cancelling out $a$ and by dividing the numerator and denominator by the fraction $\dfrac{\sin \phi}{\sin \gamma}$: $$\sin (OBC)=\dfrac{\sin \beta}{\sqrt{1+\bigg(\dfrac{\sin \gamma}{2\,\sin \phi \,\cos\alpha}\,\bigg)^2-\;2\cdot \dfrac{\sin \gamma}{2\,\sin \phi\, \cos\alpha}\,\cos \beta}}=$$ $\qquad\qquad\qquad\qquad$ and now completing the square in the denominator $$=\dfrac{\sin \beta}{\sqrt{\bigg(\dfrac{\sin \gamma}{2\,\sin \phi \,\cos\alpha}\,\bigg)^2-\;2\cdot \dfrac{\sin \gamma}{2\,\sin \phi\, \cos\alpha}\,\cos \beta \, + \, \cos^2{\beta} - \, \cos^2{\beta} \, + \, 1}}=$$ $\qquad\qquad\qquad\qquad\qquad\qquad$ remembering that $1-\cos^2 \beta=\sin^2 \beta$ $$=\dfrac{\sin \beta}{\sqrt{\bigg(\dfrac{\sin \gamma}{2\,\sin \phi \,\cos\alpha}-\cos{\beta} \bigg)^2+\;\sin^2\beta}}=$$ $\;$and then dividing the numerator and denominator by $\,\sin \beta\;$ we derive the final version of the formula $$=\dfrac{1}{\sqrt{\bigg(\dfrac{\sin \gamma}{2\,\sin \beta \,\sin \phi \,\cos\alpha}-\cot{\beta} \bigg)^2+\;1}}=$$ $\qquad\qquad\qquad\qquad$ or $$\angle OBC= \sin^{-1}\dfrac{1}{\sqrt{\bigg(\dfrac{\sin \gamma}{2\,\sin \beta \,\sin \phi \,\cos\alpha}-\cot{\beta} \bigg)^2+\;1}}$$ Plugging in the angle values we once again obtain the same approximate result: $$\angle OBC\approx\sin^{-1}\frac{1}{\sqrt{632.16}}\approx\sin^{-1}{0.039773}\approx2.279^\circ\,.$$ $\angle ACB=\alpha\qquad(\approx 6.6670^\circ)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ $\angle BCO=\beta\qquad(\approx 0.5788^\circ)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ $\angle AOC=\gamma\qquad(\approx 169.8570^\circ)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ $\angle OAC=\phi\qquad(\approx 4.0548^\circ)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$
This radical in fact can be further simplified using the formula $$\sin {(\cot^{-1}{x})}=\frac{1}{\sqrt{x^2+1}}:$$ $\qquad\qquad\qquad\qquad\qquad$ here $\;x=\dfrac{\sin \gamma}{2\,\sin \beta \,\sin \phi \,\cos\alpha}-\cot{\beta}$ $$\angle OBC=\sin^{-1}{\dfrac{1}{\sqrt{\bigg(\dfrac{\sin \gamma}{2\,\sin \beta \,\sin \phi \,\cos\alpha}-\cot{\beta} \bigg)^2+\;1}}}=$$ $$\sin^{-1}{\frac{1}{\sqrt{x^2+1}}}=\sin^{-1}{(\sin{(\cot^{-1}{x})})}=\cot^{-1}{x}=\cot^{-1}{\bigg(\frac{\sin \gamma}{2\,\sin \beta \,\sin \phi \,\cos\alpha}-\cot{\beta}\bigg)}$$
This gives the result even faster. You get the same number (can check with a calculator) $\approx2.279^\circ\,$ I've just turned the solution in my head through Ceva's theorem and it went without the radical (square root) i.e. you get the cotangent from a trigonometric equation resulting through the use of Ceva's theorem. This answer needs to be expanded to encompass Ceva's theorem, I guess. There's the beauty in all these interconnections. That's all. I need to add the second solution through Ceva's theorem. $$\quad$$ $\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$* * * * * * * * * * * *$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ $$\quad$$ Alternative solution using Ceva's theorem. Let's write Ceva's theorem in trigonometric form: $$\frac{\sin(ABO)}{\sin(OBC)}\,\frac{\sin(OAC)}{\sin(BAO)}\,\frac{\sin(BCO)}{\sin(OCA)}=1$$ $\qquad\qquad\qquad\qquad\qquad\qquad\angle ACB=\alpha\qquad(\approx 6.6670^\circ)\qquad\qquad\qquad\qquad$ $\qquad\qquad\qquad\qquad\qquad\qquad\angle BCO=\beta\qquad(\approx 0.5788^\circ)\qquad\qquad\qquad\qquad$ $\qquad\qquad\qquad\qquad\qquad\qquad\angle AOC=\gamma\qquad(\approx 169.8570^\circ)\qquad\qquad\qquad\qquad$ $\qquad\qquad\qquad\qquad\qquad\qquad\angle OAC=\phi\qquad(\approx 4.0548^\circ)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$
We use the same denotations here but there are "more" angles now. Just like was shown in the very beginning most of these angles are obvious. $\;\angle OAB=\alpha-\phi\,,\;\angle OCA=\alpha-\beta\,,\;$and denoting $\;\angle OBC\;$ (the angle we have to find) as $\;x\;$ it follows that $\;\angle ABO+\angle OBC=180-2\alpha\;$ or $\;\angle ABO=180-2\alpha-x\,.$ Now we re-write Ceva's theorem as follows $$\frac{\sin(180-2\alpha-x)}{\sin x}\,\frac{\sin \phi}{\sin(\alpha-\phi)}\,\frac{\sin \beta}{\sin(\alpha-\beta)}=1$$ Now we simplify the first factor of this equation: $$\frac{\sin(180-2\alpha-x)}{\sin x}=\frac{\sin(2\alpha+x)}{\sin x}=\frac{\sin(2\alpha)\cos x + \cos(2\alpha)\sin x}{\sin x}=\cot x \sin (2\alpha)+\cos (2\alpha)$$ So we have $$[\cot x\,\sin (2\alpha)+\cos (2\alpha)]\,\frac{\sin \phi}{\sin(\alpha-\phi)}\,\frac{\sin \beta}{\sin(\alpha-\beta)}=1$$ $$\cot x\,\sin (2\alpha)+\cos (2\alpha)=\,\frac{\sin(\alpha-\phi)\,\sin(\alpha-\beta)}{\sin \phi \,\sin \beta}$$ $$\cot x =\,\frac{\sin(\alpha-\phi)\,\sin(\alpha-\beta)}{\sin(2\alpha) \sin \phi \,\sin \beta}-\cot{(2\alpha)}$$ $$x =\,\cot^{-1}{\bigg(\,\frac{\sin(\alpha-\phi)\,\sin(\alpha-\beta)}{\sin(2\alpha) \sin \phi \,\sin \beta}-\cot{(2\alpha)}\bigg)}$$ Once again by plugging the numbers we get this value: $$x\approx\cot^{-1}{25.123}=\tan^{-1}{\frac{1}{25.123}}\approx 2.279^\circ.$$ Using Ceva's theorem gives us the solution through the inverse cotangent after solving the trivial trigonometric equation. So we don't have the radical here as opposed to when we used the law of cosines.