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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

  1. 30°

  2. 60°

  3. 45°

  4. 90°

Circles

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Answer

The circle with centre O and PQ as tangent is shown in the figure below:

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 (a) 30° (b) 60° (c) 45° (d) 90°. Circles, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

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 = 90°2\dfrac{90°}{2}
⇒ ∠QOP = 45°.

Hence, Option 3 is the correct option.

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