A curious geometry problem: Find the $\angle OBC$ 
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 .
 A: 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.

