smc

Since $△ABC$ is an isosceles triangle, $\angle ABC=\angle ACB=70^{\circ }$. Let $\angle OBC=\angle OCA=x$. Since $\angle ACB=70$, $\angle OCB=70-x$. The angle of $△OBC$ add up to $180$, so $\angle BOC=180-(x+70-x)=\boxed{110^{\circ }}$.