Question

PQ is a tangent drawn from a point P to a circle with center O and QOR is a diameter of the circle such that $\angle$POR $={120}^{\circ }$, then $\angle$OPQ is ${30}^{\circ }$.

• A. 60°
• B. 45°

Answer 1

The correct option is A

60°

Given that $\angle POR={120}^{\circ }$

We know that $\angle$OQP is ${90}^{\circ }$

Using external angle theorem, $\angle POR=\angle OQP+\angle OPQ$

$\Rightarrow {120}^{\circ }={90}^{\circ }+\angle OPQ$

$\Rightarrow \angle OPQ={120}^{\circ }–{90}^{\circ }$

$\Rightarrow \angle OPQ={30}^{\circ }$