In △AOP and △BOP

AP = BP

[Tangents drawn from an external point are equal]

OP = OP [Common]

OA = OB

[Radii of same circle]

△AOP ≅ △BOP

[By Side - Side - Side criterion]

∠APO = ∠BPO

[Corresponding parts of congruent triangles are congruent]

∠APB = ∠APO + ∠BPO

80 = ∠APO + ∠APO

2∠APO = 80

∠APO = 40°

In △AOP

∠APO + ∠AOP + ∠OAP = 180°

[By angle sum property]

40° + ∠AOP + 90° = 180°

[ ∠OAP = 90° as OA ⏊ AP because Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∠AOP = 50°