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Answer :
Since PA and PB are tangents to a circle with centre O,
\(\therefore OA \ ⊥ \ AP \), and \( OB \ ⊥ \ BP \)
\( ∠ \ OAP \ = \ 90° \) and \( ∠ \ OBP=90° \)
In the quadrilateral PAOB, we have
\( \Rightarrow ∠ \ ABP \ + \ ∠ \ PAO \ + \ ∠ \ AOB \ + \ ∠ \ OBP \ = \ 360° \)
\( \Rightarrow \ ∠ \ 80° \ + \ 90° \ + \ ∠ \ AOB \ + \ 90° \ = \ 360° \)
\( \Rightarrow \ 260° \ + \ ∠ \ AOB \ = \ 360° \)
\( \Rightarrow \ ∠ \ AOB \ = \ 360° \ - \ 260° \ = \ 100° \)
In \( ∆ \ OAP \) and \( OBP \), we have
\( OP \ = \ OP \) [Common]
\( OA \ = \ OB \) [Radii]
\( ∠ \ OAP \ = \ ∠ \ OBP \) [Each = 90°]
\(\therefore \) \( ∆ \ OAP \ \cong \ ∆ \ OBP \) ( By SAS Criterion)
\(\Rightarrow \ ∠ \ POA \ = \ ∠ \ POB \) [C.P.C.T.]
\(\therefore \) \( ∠ \ POA \ = \ \frac{1}{2} \ ∠ \ AOB \ \)
\( = \ \frac{1}{2} \ × \ 100° \ = \ 50° \)
∴ The (A) is the correct option.