#### 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°

clickto share

#### Solution

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

`anglePOA = 1/2 angleAOB = (100^@)/2 = 50^@`

Hence, alternative (A) is correct.

Is there an error in this question or solution?

#### Related QuestionsVIEW ALL [18]

#### Reference Material

Solution for 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 concept: null - Number of Tangents from a Point on a Circle. For the course CBSE