(b) $90^\circ$
Since PQ is a tangent at point P, by Theorem 10.1, OP ⊥ PQ, so ∠OPQ = 90°. In △OPQ, x + y + ∠OQP = 180°, but since ∠OPQ = x = 90° − y (as OP ⊥ PQ means x + y = 90°). Thus, x + y = 90°.
By Theorem 10.1, the radius OP is perpendicular to the tangent PQ at the point of contact, so ∠OPQ = 90°. Since ∠OPQ = x and ∠POQ = y, and both angles are part of triangle OPQ where the right angle is at P, we get x + y = 90°. Examiner expects you to recall Theorem 10.1 directly to justify the answer.