Q1. [1] § 10.1 Introduction § 10.2 Tangent to a Circle § 10.4 Summary
In the given figure, PA is a tangent from an external point P to a circle with centre O. If $\angle POB = 125°$, then $\angle APO$ is equal to :
- (a) $25°$
- (b) $65°$
- (c) $90°$
- (d) $35°$
Previously asked in CBSE board exam
2026 30/1/1 Q13
Generated by claude-sonnet-4-6 · 2026-06-15 10:25 · grounding rag
Model Answer
(d) 35°
Since PA is a tangent, ∠OAP = 90°. ∠AOP = 180° − 125° = 55° (angles on a straight line). In △OAP: ∠APO = 180° − 90° − 55° = 35°.
Explanation
Key steps: (1) Radius OA ⊥ tangent PA (Theorem 10.1), so ∠OAP = 90°. (2) ∠AOP + ∠POB = 180° (linear pair), giving ∠AOP = 55°. (3) Angle sum in △OAP gives ∠APO = 35°. Examiners expect you to state the tangent-radius perpendicularity and use the linear pair/angle sum property clearly.
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