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First off, I would like to apologise for the low quality image.
As $\angle OBC=\angle OCB,$ we conclude that $\triangle OBC$ is isosceles, thus $OB=OC=2$.
Then we use the fact that $\sin \theta= \frac {\text{opposite}}{\text{hypotenuse}}.$ Thus $\angle ACO= 30^\circ.$
As $AH=CH,$ we conclude that $\triangle OAC$ is isosceles, thus $\angle ACO=\angle CAO=30^\circ$.
Using the fact that the angles in a triangle sum to $180^\circ,$ we find that $\angle BOC=140^\circ$ and $\angle AOC=120^\circ.$
$\therefore \angle ABO=100^\circ$
As $\angle OAC=30^\circ$ and $OH=1, OA=2.$
As both $OB$ and $OA$ are equal to $2,$ $\triangle OAB$ is isosceles and $\angle ABO=\angle BAO=x^\circ$
$x+x+100=180 \implies x=40^\circ$
$\therefore\angle ABO=40^\circ$