Question

If O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50 degree with PQ, then find angle POQ.

Answer 1

OPR=90( radius is always perpendicular to tangent)
RPQ+QPO=90
QPO=90-50
QPO=40
OP=OQ (radii)
OPQ=OQP=40(angles opp to equal sides)
In triangle OPQ
OPQ+OQP+POQ=180(ASP)
POQ=180-80
POQ=100

Answer 2

Given : O is the center of the circle ,PQ is chord & the tangent PR at P makes an angle 50° with PQ

To Find : ∠POQ

Solution:

PR is tangent

Hence ∠OPR = 90°

tangent PR at P makes an angle 50° with PQ

=> ∠QPR =  50°

∠OPR = ∠OPQ + ∠QPR

=> 90° = ∠OPQ +   50°

=> ∠OPQ  = 40°

∠OQP = ∠OPQ   ∵ OP = OQ  ( Radius)

=> ∠OQP = 40°

in  ΔOPQ  sum of angles  of triangle = 180°

=> ∠OQP + ∠OPQ + ∠POQ = 180°

=> 40° + 40° +  ∠POQ = 180°

=> ∠POQ = 100°

∠POQ = 100°

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