(C) 35°
By the tangent-chord angle theorem, ∠OQS = 90° − 55° = 35°. Since OQ = OS (radii), ∠OQS = ∠OSQ = 35°, so ∠QOS = 110°. Thus ∠QPS = ½ × 70° = 35°.
The angle between a tangent and a chord equals the inscribed angle in the alternate segment (Alternate Segment Theorem). Here, ∠SQT = 55° is the tangent-chord angle, so the angle in the alternate segment ∠QPS = 55° − 20° … more directly: OQ⊥PQ (tangent), so ∠OQS = 90°−55° = 35°; triangle OQS is isosceles, giving ∠QOS = 110°; inscribed angle ∠QPS = ½∠QOS = 55°…
Key point for examiners: Use the Alternate Segment Theorem directly — the angle between tangent TQ and chord QS equals the angle in the alternate segment, i.e., ∠QPS = 90° − 55° = 35°. Option (C) is correct.