Question

PQ is a tangent to a circle with centre O at the point P. If $△$OPQ is an isosceles triangle such that OP = PQ, then find the measure of $\angle$OQP.

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is B

${45}^{\circ }$

In $△$OPQ, OP = PQ
$\therefore \angle POQ=\angle PQO$

$\angle OPQ={90}^{\circ }$
(Radius is perpendicular to tangent at the point of contact)

We have,

$\angle POQ+\angle PQO+\angle OPQ={180}^{\circ }$
[Angle sum property of a triangle]

$2\angle PQO+{90}^{\circ }={180}^{\circ }$

$\angle PQO=\frac{{90}^{\circ }}{2}={45}^{\circ }$