Mathematics
PQ is a tangent to a circle at point P. Centre of circle is O. If △OPQ is an isosceles triangle, then ∠QOP is equal to
30°
60°
45°
90°
Circles
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Answer
The circle with centre O and PQ as tangent is shown in the figure below:
We know that,
OP ⊥ PQ (∵ tangent through a point and radius from that point are perpendicular to each other)
Given, △OPQ is an isosceles triangle.
Since, ∠OPQ = 90° hence, the other two angles will be equal to each other.
∴ ∠QOP = ∠OQP
We know that sum of angles in a triangle = 180°.
In △OPQ,
⇒ ∠OPQ + ∠QOP + ∠OQP = 180°
⇒ ∠OPQ + ∠QOP + ∠QOP = 180°
⇒ 90° + 2∠QOP = 180°
⇒ 2∠QOP = 180° - 90
⇒ ∠QOP =
⇒ ∠QOP = 45°.
Hence, Option 3 is the correct option.
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