Find poq of tangents drawn to the circle, Mathematics

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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)

2314_circle5.png

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

In ΔAPQ

∠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

= 180o-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

∠POQ =125o

∴∠POQ =125o


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