Answer :

Given: PA and PB are tangents.

Therefore, the radius drawn to these tangents will be perpendicular to the tangents.

Thus, OA ⊥ PA and OB ⊥ PB

∠OBP = 90°

∠OAP = 90°

In quadrilateral AOBP,

Sum of all interior angles = 360°

∠OAP + ∠APB +∠PBO + ∠BOA = 360° 90° + 80° +90° + ∠BOA = 360°

∠BOA = 100°

In ΔOPB and ΔOPA,

AP = BP (Tangents from a point)

OA = OB (Radii of the circle)

OP = OP (Common side)

Therefore, ΔOPB ≅ ΔOPA (SSS congruence criterion)

And thus, ∠POB = ∠POA

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