Q. 13.7( 3 Votes )

# In the given figu A. 130°

B. 100°

C. 90°

D. 75°

Answer :

In the given figure PT is a tangent to circle ∴ we have

∠OPT = 90°

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

∠OPQ + ∠QPT = 90°

∠OPQ + 50° = 90°

∠OPQ = 40°

Now, In △POQ

OP = OQ

∠PQO = ∠QPO = 40°

[Angles opposite to equal sides are equal]

Now,

∠ PQO + ∠QPO + ∠ POQ = 180° [

By angle sum property of triangle]

40° + 40° + ∠POQ = 180°

∠POQ = 100°

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