PQ is a tangent to a circle with centre O at the point P.


PQ is a tangent to a circle with centre O at the point P. If Δ OPQ is an isosceles triangle, then ∠OQP is equal to

(a)  30°

(b) 45°

(c) 60°

(d) 90°


Let us first put the given data in the form of a diagram.

We know that the radius of a circle will always be perpendicular to the tangent at the point of contact. Therefore, OP is perpendicular to QP. Therefore,

$\angle O P Q=90^{\circ}$

The side opposite tois OQ.OQ will be the longest side of the triangle. So, in the isosceles right triangle ΔOPQ,


And the angles opposite to these two sides will also be equal. Therefore,

We know that sum of all angles of a triangle will always be equal to 180°. Therefore,

$\angle O Q P+\angle P O Q+\angle O P Q=180^{\circ}$

$\angle O Q P+\angle P O Q+\angle O P Q=180^{\circ}$

$90^{\circ}+2 \angle O Q P=180^{\circ}$

$2 \angle O Q P=90^{\circ}$

$\angle O Q P=45^{\circ}$

Therefore, the correct answer to this question is choice (b).


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