Answer :

Given: PT is a tangent to a circle with center O and PQ is a chord of the circle such that ∠QPT = 70°

To Find: ∠POQ = ?

Now,

∠OPT = 90°

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∠OPQ + ∠QPT = 90°

∠OPQ + 70° = 90°

∠OPQ = 20°

Also,

OP = OQ [Radii of same circle]

∠OQP = ∠OPQ = 20°

[Angles opposite to equal sides are equal]

In △OPQ By Angle sum property of triangles,

∠OPQ + ∠OQP + ∠POQ = 180°

20° + 20° + ∠POQ = 180°

∠POQ = 140°

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