**Solution:**

OA and OB are radii of the circle to the tangents PA and PB respectively.

∴ OA ⊥ PA and,

∴ OB ⊥ PB

∠OBP = ∠OAP = 90°

In **quadrilateral **AOBP,

**Sum of all interior angles = 360°**

∠AOB + ∠OBP + ∠OAP + ∠APB = 360°

⇒ ∠AOB + 90° + 90° + 80° = 360°

⇒ ∠AOB = 100°

Now, In ΔOPB and ΔOPA,

AP = BP (Tangents from a point are equal)

OA = OB (Radii of the circle)

OP = OP (Common side)

∴ ΔOPB ≅ ΔOPA (by SSS congruence condition)

Thus ∠POB = ∠POA

∠AOB = ∠POB + ∠POA

⇒ 2 ∠POA = ∠AOB

⇒ ∠POA = 100°/2 = 50°

∠POA is equal to option (A) 50°