Question

If tangents PA and PB from a point P to a circle with centre O are inclined to each other an angle of 80, then ∠POA is equal to

(A) 50 (B) 60

(C) 70 (D) 80

Answer 1

It is given that PA and PB are tangents.

Therefore, the radius drawn to these tangents will be perpendicular to the tangents.

Thus, OA ⊥ PA and OB ⊥ PB

∠OBP = 90º

∠OAP = 90º

In AOBP,

Sum of all interior angles = 360

∠OAP + ∠APB +∠PBO + ∠BOA = 360

90 + 80 +90º +BOA = 360

∠BOA = 100

In ΔOPB and ΔOPA,

AP = BP (Tangents from a point)

OA = OB (Radii of the circle)

OP = OP (Common side)

Therefore, ΔOPB ≅ ΔOPA (SSS congruence criterion)

A ↔ B, P ↔ P, O ↔ O

And thus, ∠POB = ∠POA

Hence, alternative (A) is correct.