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

Question:

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°

Solution:

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,

OP = PQ

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).