Mathematics — CBSE Class 10 board question

Given: PA and PB are tangents from external point P to a circle with centre O.

To Prove: $\angle APB = 2\angle OAB$

Proof:

Let $\angle APB = \theta$

Since PA = PB (tangents from an external point are equal), $\triangle PAB$ is isosceles.

$$\angle PAB = \angle PBA = \frac{1}{2}(180° - \theta) = 90° - \frac{\theta}{2}$$

By Theorem 10.1, $\angle OAP = 90°$ (radius $\perp$ tangent)

$$\therefore \angle OAB = \angle OAP - \angle PAB = 90° - \left(90° - \frac{\theta}{2}\right) = \frac{\theta}{2}$$

$$\therefore \angle APB = \theta = 2 \times \frac{\theta}{2} = 2\angle OAB \quad \textbf{(Proved)}$$

Source: Chapter 10, Section 10.3 (Example 2)

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• This proof directly mirrors Example 2 of the textbook (with different letter labels). Examiners expect this exact approach.
• Key facts used: (1) tangents from external point are equal → isosceles triangle, (2) radius ⊥ tangent → $\angle OAP = 90°$.
• Always state "Let $\angle APB = \theta$" to set up algebra cleanly — it avoids confusion and earns method marks.
• Write "Proved" or the QED box at the end; examiners look for a clear conclusion.