If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠POA is equal to.

The correct option is A 50o
Given: ∠APB = 80°
As we know that sum of ∠APB + ∠AOB = 180° (Using Property)
∠AOB = 100°
ΔAOP ≅ ΔBOP by SSS congruency criteria.
AO = BO (Radius of circle)
PO = PO (common side of triangle)
AP = BP (tangents from an external point)
∠AOP = ∠BOP (CPCT )
∠POA + ∠POB = ∠AOB = 100°
∠POA = 50°