3 Tutor System
Starting just at 265/hour

# 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.