If you find any point $C$ such that $\angle ACB = £/2$, then the $AB$ arc of the circumcircle of $\triangle ABC$ will work.
You can do this by finding any two angles that add to $180^\circ - £/2$, and drawing one of them at $A$ and the other one at $B$, both with $AB$ as one of the angle legs, and let $C$ be where the two new angle legs intersect.