Q1. [3] § 10.1 Introduction § 10.2 Tangent to a Circle § 10.4 Summary
In the adjoining figure, TP and TQ are tangents drawn to a circle with centre O. If $\angle OPQ = 15°$ and $\angle PTQ = \theta$, then find the value of $\sin 2\theta$.
Previously asked in CBSE board exam
2025 30/6/1 Q31
Generated by claude-sonnet-4-6 · 2026-06-15 10:27 · grounding rag
Model Answer
Given: TP and TQ are tangents from external point T to circle with centre O. ∠OPQ = 15°, ∠PTQ = θ.
From Example 2 (Theorem): $\angle PTQ = 2\angle OPQ$
$$\theta = 2 \times 15° = 30°$$
Therefore:
$$\sin 2\theta = \sin(2 \times 30°) = \sin 60° = \dfrac{\sqrt{3}}{2}$$
$$\boxed{\sin 2\theta = \dfrac{\sqrt{3}}{2}}$$
Source: Chapter 10, Example 2
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Explanation
- The key result used here is from Example 2 of the chapter: $\angle PTQ = 2\angle OPQ$. This is derived using the fact that TP = TQ (equal tangents), making △TPQ isosceles, and ∠OPT = 90° (radius ⊥ tangent).
- Once θ = 30°, compute sin 2θ = sin 60° directly.
- Examiners award marks for: identifying the relationship θ = 2 × 15°, finding θ = 30°, and correctly evaluating sin 60°. Show each step clearly.
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