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In the given figure, O is the center of a circle, PQ is a chord and the tangent PT at P makes an angle of 50° with PQ. Then, ∠POQ = ?
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]
∠ PQO + ∠QPO + ∠ POQ = 180° [
By angle sum property of triangle]
40° + 40° + ∠POQ = 180°
∠POQ = 100°