jmc

The angle $\theta$ is the angle formed by two perpendiculars drawn to $BO$, one on the plane determined by $OAB$ and the other by $OBC$. Let the perpendiculars from $A$ and $C$ to $\overline{OB}$ meet $\overline{OB}$ at $P.$ Without loss of generality, let $AP=1.$ It follows that $△OPA$ is a $45-45-90$ right triangle, so $OP=AP=1,$ $OB=OA=\sqrt{2},$ and $AB=\sqrt{4-2\sqrt{2}}.$ Therefore, $AC=\sqrt{8-4\sqrt{2}}.$ From the Law of Cosines, $A{C}^2=A{P}^2+P{C}^2-2(AP)(PC)\cos \theta ,$ so $$8-4\sqrt{2}=1+1-2\cos \theta ⟹\cos \theta =-3+2\sqrt{2}=-3+\sqrt{8}.$$ Thus $m+n=\boxed{5}$.