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

Given: PA and PB are two tangents of the circle.
We know that the radius is perpendicular to tangent. Therefore, OA ⊥PA and OB PB.
⇒ ∠OBP = 90° and ∠OAP = 90°
In quadrilaterral AOBP, ∠OAP + ∠APB + ∠PBO + ∠BOA = 360°
⇒ 90° + 80° + 90° + ∠BOA = 360°
⇒ ∠BOA = 360° – 260° = 100°
In ΔOPB and ΔOPA,
AP = BP [Tangents drawn from same external point]
OA = OB [Radii]
OP = OP [Common]
Therefore, ΔOPB ≅ ΔOPA [SSS Congruency rule]
Hence, ∠POB = ∠POA
∠POA = 1/2 ∠AOB = 1/2(100°) = 50°. Hence, the option (A) is correct.