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 $\angle AOB = 130°$, then $\angle APB$ is equal to :
- A 130°
- B 50°
- C 120°
- D 90°
Previously asked in CBSE board exam
2026 30/2/1 Q14
Generated by claude-sonnet-4-6 · 2026-06-15 10:25 · grounding rag
Model Answer
Option B: 50°
Since OA ⊥ PA and OB ⊥ PB (radius ⊥ tangent), in quadrilateral OAPB:
∠APB = 360° − 90° − 90° − 130° = 50°
Explanation
In quadrilateral OAPB, the four angles sum to 360°. Angles OAP and OBP are each 90° (Theorem 10.1). So ∠APB = 360° − 130° − 90° − 90° = 50°. This is a standard application of the tangent-radius perpendicularity theorem. Note that ∠APB + ∠AOB = 180° (they are supplementary), which is a quick shortcut: 180° − 130° = 50°.
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