How to find angle $x$?

Question

In the figure above, point $O$ is the center of the circle and $OC=AC=AB$. What is the value of $x$?

$\text m\angle OAC=x\unicode{176}$.

$\text m\angle OAB=\text m \angle ABO$.

What else I need to deduce to solve for $x$?

obviously $\text m\angle AOC+\text m\angle OBA+\text m \angle BAO=180\unicode{176}$.

$\text m \angle ACB+\text m \angle CBA+\text m \angle BAC=180\unicode{176}$.

$\text m \angle OAC+x+\text m \angle CAO=180\unicode{176}$.

Answer 1

Steps:
From $\triangle OCA$, since we know $OC=AC \implies \angle COA = \angle CAO \quad (=x^{\circ})$.
From $\triangle OBA$, since we know that both $OB$ and $OA$ are equal to radius of the given circle $\implies \angle OAB = \angle OBA \quad (=y^{\circ})$.
From $\triangle CAB$, since we know $AC=AB \implies \angle ACB = \angle ABC \quad (=y^{\circ})$.

Finally, since the sum of all interior angles in both triangles $\triangle OCA$ and $\triangle CAB$ is equal to $180^{\circ}$, we can easily find $\boxed{x=36^{\circ}}.$

Answer 2

Basically, you have a triangle with one size $s$ and two sizes $2s$.
This is a simple usage of the cosine-rule.