Because AOB is the diameter of the circle, \[ \therefore \angle ACB = 90^\circ \] So, \[ \angle BAC = 90^\circ - \angle ABC = 90^\circ - \angle OBC = 90^\circ - 60^\circ = 30^\circ \] \[ \therefore \angle OAC = 30^\circ \] Again, in triangle OCA, since OC = OA = radius of the circle, \[ \therefore \angle OCA = \angle OAC = 30^\circ \]