In the given figure, PA and PB are tangents to a circle centred at ...

Question

In the given figure, PA and PB are tangents to a circle centred at O. If $\angle$ OAB = $15^\circ$, then $\angle$ APB equals :

*(Figure : two tangents PA and PB drawn from external point P, O is the centre, AB is the chord of contact.)*

Accepted Answer

**Step 1 — Use the tangent–radius property.**
$OA \perp PA$, so $\angle OAP = 90^\circ$.

**Step 2 — Find $\angle PAB$.**
$$\angle PAB = \angle OAP - \angle OAB = 90^\circ - 15^\circ = 75^\circ$$

**Step 3 — Use equal tangents.**
$PA = PB$ (tangents from an external point are equal)
$\Rightarrow 	riangle PAB$ is isosceles $\Rightarrow \angle PBA = \angle PAB = 75^\circ$

**Step 4 — Apply the angle sum property in $	riangle PAB$.**
$$\angle APB = 180^\circ - (75^\circ + 75^\circ) = 180^\circ - 150^\circ = 30^\circ$$

**Answer : (A) $30^\circ$**

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