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,
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).
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