In figure, if 0 is the center of a circle, PQ is a chord and the tangent PR at P makes an angle of 50° with PQ, then ∠POQ is equal to

Given : OP is a radius and PR is a tangent in a circle with center O with ∠RPQ = 50°

To find : ∠POQ

Now, OP ⏊ PR [ As tangent to at any point on the circle is perpendicular to the radius through point of contact]

∠OPR = 90°

∠OPQ + ∠RPQ = 90°

∠OPQ + 50° = 90°

∠OPQ = 40°

In △POQ

OP = OQ [radii of same circle]

∠OQP = ∠OPQ = 40° [ angles opposite to equal sides are equal] [1]

In △OPQ By angle sum property of a triangle

∠OPQ + ∠OPQ + ∠POQ = 180°

40° + 40° + ∠POQ = 180°

∠POQ = 100°

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