In Given Figure,

PA = PB…[1]

[Tangents drawn from an external point are equal]

In △AOP and △BOP

PA = PB [By 1]

OP = OP [Common]

OA = OB

[radii of same circle]

△AOP ≅△BOP

[By Side - Side - Side Criterion]

∠OPA = ∠OPB

[Corresponding parts of congruent triangles are congruent]

Now,

∠APB = 80° [Given]

∠OPA + ∠OPB = 80°

∠OPA + ∠OPA = 80°

2 ∠OPA = 80°

∠OPA = 40°

In △AOP,

∠OAP = 90°

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

And

∠OAP + ∠OPA + ∠AOP = 180°

90° + 40° + ∠AOP = 180°

∠AOP = 50°