Question

In the figure PA and PB are tangents to the circle C(O, r) from an external point P. If angle between in these tangents is 70°, then ∠POA is equal to.

(A) 60°

(B) 110°

(C) 55°

(D) 90°.

Answer 1

Answer is (C) 55°

PA and PB are tangents to a circle from point P and OA and OB are radius of circle. So AP ⊥ OA and PB ⊥ OB.

∴ ∠OAP = 90°

and ∠OBP = 90°

Now ∠AOB + ∠APB = 180°

⇒ ∠AOB + 70° = 180°

⇒ ∠AOB = 180° – 70°

∴ ∠AOB = 110°

∵ line OP is bisect of ∠AOB

∴ ∠POA = \(\frac { \angle AOB }{ 2 } \) = \(\frac { { 110 }^{ \circ } }{ 2 } \) = 55°

∴ ∠POA = 55°