Q1. [1] § 10.1 Introduction § 10.2 Tangent to a Circle
In the given figure, PQ is a tangent to the circle with centre O. If $\angle OPQ = x$, $\angle POQ = y$, then $x + y$ is :
- (a) $45^\circ$
- (b) $90^\circ$
- (c) $60^\circ$
- (d) $180^\circ$
Previously asked in CBSE board exam
2023 30/4/1 Q14
Generated by claude-sonnet-4-6 · 2026-06-15 10:23 · grounding rag
Model Answer
(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°.
Explanation
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.
If a question refers to an image, map, graph or diagram that is not shown here, find the actual CBSE question paper on the CBSE website:
cbse.gov.in.