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 $\angle$ AOB = $130^\circ$, then $\angle$ APB is equal to :

*(Figure : two tangents PA and PB from external point P to a circle with centre O, $\angle AOB = 130^\circ$.)*

Accepted Answer

**Step 1 — Use the tangent–radius property.**
$$\angle OAP = \angle OBP = 90^\circ$$

**Step 2 — Apply the angle sum property of quadrilateral OAPB.**
$$\angle AOB + \angle OAP + \angle APB + \angle OBP = 360^\circ$$

**Step 3 — Substitute the known values.**
$$130^\circ + 90^\circ + \angle APB + 90^\circ = 360^\circ$$

**Step 4 — Solve.**
$$\angle APB = 360^\circ - 310^\circ = 50^\circ$$

*(Shortcut : $\angle AOB + \angle APB = 180^\circ$, so $\angle APB = 180^\circ - 130^\circ = 50^\circ$.)*

**Answer : (B) $50^\circ$**