Q1. [1] § 10.1 Introduction § 10.2 Tangent to a Circle § 10.4 Summary
In the given figure, PA and PB are tangents to a circle centred at O. If $\angle OAB = 15°$, then $\angle APB$ equals :
- A 30°
- B 15°
- C 45°
- D 10°
Previously asked in CBSE board exam
2026 30/2/1 Q13
Generated by claude-sonnet-4-6 · 2026-06-15 10:25 · grounding rag
Model Answer
Answer: A (30°)
Since OA ⊥ PA (radius ⊥ tangent), ∠OAP = 90°, so ∠OAB = 15° means ∠PAB = 90° − 15° = 75°. By symmetry ∠PBA = 75°, so ∠APB = 180° − 75° − 75° = 30°.
Explanation
Key facts used: (1) Radius ⊥ tangent at point of contact (Theorem 10.1), so ∠OAP = 90°. (2) PA = PB (Theorem 10.2), so △PAB is isosceles. Since ∠OAB = 15°, ∠PAB = 90° − 15° = 75°; by symmetry ∠PBA = 75°; hence ∠APB = 180° − 150° = 30°. Examiners expect you to state both theorems clearly.
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