If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then \( ∠ \ POA \) is equal to
(A) \( 50° \)
(B) \( 60° \)
(C) \( 70° \)
(D) \( 80° \)

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.