choose the correct option and give justification.
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°
OA and OB are radii of the circle to the tangents PA and PB respectively.
∴ OA ⊥ PA and,
∴ OB ⊥ PB
∠OBP = ∠OAP = 90°
In quadrilateral AOBP,
Sum of all interior angles = 360°
∠AOB + ∠OBP + ∠OAP + ∠APB = 360°
⇒ ∠AOB + 90° + 90° + 80° = 360°
⇒ ∠AOB = 100°
Now,
In ΔOPB and ΔOPA,
AP = BP (Tangents from a point are equal)
OA = OB (Radii of the circle)
OP = OP (Common side)
∴ ΔOPB ≅ ΔOPA (by SSS congruence condition)
Thus ∠POB = ∠POA
∠AOB = ∠POB + ∠POA
⇒ 2 ∠POA = ∠AOB
⇒ ∠POA = 100°/2 = 50°
∠POA is equal to option (A) 50°
