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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: 125o)
Ans: Given ∠PAB = 125o
To find : - ∠POQ = ?
Construction : - Join PQ
Proof : - ∠PAB + ∠PAQ = 180o (Linear pair)
∠PAQ + 125o = 180o
∠PAQ = 180o - 125o
∠PAQ = 55o
Since the length of tangent from an external point to a circle are equal.
PA = QA
∴ From ΔPAQ
∠APQ = ∠AQP
∠APQ + ∠AQP + ∠PAQ = 180o (angle sum property)
∠APQ + ∠AQP + 55o = 180o
2∠APQ = 180o - 55o (Θ ∠APQ = ∠AQP)
∠APQ = 125/2
∴∠APQ =∠AQP = 125/2
OQ and OP are radii
QA and PA are tangents
∴∠OQA = 90o
& ∠OPA = 90o
∠OPQ + ∠QPA = ∠OPA = 90o (Linear Pair)
∠OPQ + 125o/2 =90o
∠OPQ = 90o - 125o/2
∠OPQ = 55o/2
Similarly ∠OQP + ∠PQA = ∠OQA
∠OQP + 125o/2 =90o
∠OQP = 90o- 125o/2
∠OQP = In ΔPOQ
∠OQP + ∠OPQ + ∠POQ = 180o (angle sum property)
55o/2 + 55o/2 +∠POQ =180o
∠POQ + 110/2 =180o
∠POQ =180o - 110/2
∠POQ = 360o-110/2
∠POQ = 250o/2
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