Mathematics — Standard (041) — CBSE Class 10 board question

Given: A circle with centre O. PA and PB are two tangents drawn from external point P, where A and B are points of contact.

To Prove: ∠APB + ∠AOB = 180°

Proof:

Since OA ⊥ PA and OB ⊥ PB (radius ⊥ tangent at point of contact),

∠OAP = 90° and ∠OBP = 90°

In quadrilateral OAPB:

∠OAP + ∠APB + ∠OBP + ∠AOB = 360°

90° + ∠APB + 90° + ∠AOB = 360°

∠APB + ∠AOB = 360° − 180°

∴ ∠APB + ∠AOB = 180°

Hence, the angle between the two tangents is supplementary to the angle subtended by the line segment joining the points of contact at the centre. [Proved]

Source: Chapter 10, Exercise 10.2 Q.10; Theorem 10.1, Section 10.2

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• The key fact used is Theorem 10.1: radius ⊥ tangent at point of contact, giving two right angles.
• The proof uses the angle sum property of a quadrilateral (= 360°) applied to quadrilateral OAPB.
• Examiners expect a clear diagram description (or actual diagram), stating what is given, what to prove, and a clean algebraic step from 360°. Write "Hence Proved" or "Proved" at the end.
• This is a standard theorem — memorise the setup: quadrilateral OAPB has two right angles at A and B.