In figure, O is the centre of the Circle .AP and AQ two tangents drawn to the circle. B is a point on the tangent QA and ∠ PAB = 125 ° , Find ∠ POQ. (Ans: 125^o)

Ans:    Given ∠PAB = 125^o

To find : - ∠POQ = ?

Construction : - Join PQ

Proof : - ∠PAB + ∠PAQ = 180^o (Linear pair)

∠PAQ + 125^o = 180^o

∠PAQ = 180^o - 125^o

∠PAQ = 55^o

Since the length of tangent from an external point to a circle are equal.

PA = QA

∴ From ΔPAQ

∠APQ = ∠AQP

In ΔAPQ

∠APQ + ∠AQP + ∠PAQ = 180^o (angle sum property)

∠APQ + ∠AQP + 55o = 180^o

2∠APQ = 180^o - 55^o  (Θ ∠APQ = ∠AQP)

∠APQ = 125/2

∴∠APQ =∠AQP = 125/2

OQ and OP are radii

QA and PA are tangents

∴∠OQA = 90^o

& ∠OPA = 90^o

∠OPQ + ∠QPA = ∠OPA = 90^o (Linear Pair)

∠OPQ + 125^o/2 =90^o

∠OPQ = 90^o - 125^o/2

= 180^o-125^o/2

∠OPQ = 55^o/2

Similarly     ∠OQP + ∠PQA = ∠OQA

∠OQP + 125^o/2 =90^o

∠OQP = 90^o- 125^o/2

∠OQP = In ΔPOQ

∠OQP + ∠OPQ + ∠POQ = 180^o (angle sum property)

55^o/2 + 55^o/2 +∠POQ =180^o

∠POQ + 110/2 =180^o

∠POQ =180^o - 110/2

∠POQ = 360^o-110/2

∠POQ = 250^o/2

∠POQ =125^o

∴∠POQ =125^o