Q1. [2] § 10.1 Introduction § 10.2 Tangent to a Circle § 10.4 Summary
In the given figure, $O$ is the centre of the circle. If $\angle AOB = 145°$, then find the value of $x$.
Previously asked in CBSE board exam
2024 30/4/1 Q24
Generated by claude-sonnet-4-6 · 2026-06-15 10:23 · grounding rag
Model Answer
From the figure, OAPB is a quadrilateral where PA and PB are tangents from external point P.
Since tangent ⊥ radius:
∠OAP = ∠OBP = 90°
In quadrilateral OAPB:
∠AOB + ∠OAP + ∠APB + ∠OBP = 360°
$$145° + 90° + x + 90° = 360°$$
$$x = 360° - 325° = \mathbf{35°}$$
Source: Chapter 10, Section 10.3 (Theorem 10.1 and properties of tangents)
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Explanation
- The figure shows an external point P with two tangents PA and PB to the circle with centre O, where ∠AOB = 145° is the central angle and x = ∠APB.
- Key property used: radius ⊥ tangent at point of contact → ∠OAP = ∠OBP = 90°.
- All four angles of quadrilateral OAPB sum to 360°. Examiners expect you to state the property, set up the equation, and solve clearly.
- This is a standard result: ∠APB + ∠AOB = 180° (supplementary), so x = 180° − 145° = 35° is a quick check.
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