Given:

∠APB = 80°

Property 1: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 2: Sum of all angles of a quadrilateral = 360°.

Property 3: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

By property 1,

∠PAO = 90°

∠PBO = 90°

By property 2,

∠APB + ∠PAO + ∠PBO + ∠AOB = 360°

⇒ ∠AOB = 360° - ∠APB + ∠PAO + ∠PBO

⇒ ∠AOB = 360° - (80° + 90° + 90°)

⇒ ∠AOB = 360° - 260°

⇒ ∠AOB = 100°

Now, in ∆POA and ∆POB

OA = OB [∵ radius of circle]

PA = PB [By property 3 (tangent from P)]

OP = OP [∵ common]

∴ By SSS congruency,

∆POA ≅ ∆POB

Hence, by CPCTC

∠POA = ∠POB

Now,

∠AOB = 100°

⇒ ∠POA + ∠POB = 100° [∵∠AOB = ∠POA + ∠POB]

⇒ ∠POA + ∠POA = 100° [∵∠POA = ∠POB]

⇒ 2∠POA = 100°

⇒ ∠POA = 50°

Hence, ∠POA = 50°