In triangle ABC, \[ \angle ABC = 180^\circ - (\angle BAC + \angle BCA) = 180^\circ - (85^\circ + 55^\circ) = 40^\circ \] ∴ The central angle \(\angle AOC = 2 \times \angle ABC = 2 \times 40^\circ = 80^\circ\) Now, in triangle AOC, since AO = OC (radii of the same circle), \[ \angle OAC = \angle OCA = \frac{1}{2}(180^\circ - 80^\circ) = 50^\circ \] ∴ \(\angle OAC = 50^\circ\) (Answer)