Answer is A. 50°
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 = \(\frac{100^°}{2}\)
⇒ ∠POA = 50°
Hence, ∠POA = 50°
