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
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
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
`anglePOA = 1/2 angleAOB = (100^@)/2 = 50^@`
Hence, alternative 50° is correct.