Question

In the given figure, LMN is the tangent to the circle with centre O. If $\angle$ PMN $={60}^{\circ }$, find $\angle$MOP.

• A. ${30}^{\circ }$
• B. ${60}^{\circ }$
• C. ${90}^{\circ }$
• D. ${120}^{\circ }$

Answer 1

The correct option is D

${120}^{\circ }$

Note that $\angle OMN=\angle OMQ={90}^{\circ }(\because$A tangent at any point of a circle is perpendicular to the radius at the point of contact)

$\angle OMN={90}^{\circ },\angle PMN={60}^{\circ }⟹\angle OMP={30}^{\circ }$

We have $OP=OM$ (Radii of the same circle)
$⟹\angle OMP=\angle OPM={30}^{\circ }(\because$Base angles of isosceles triangle)

Now $\angle MOP={180}^{\circ }-(\angle OMP+\angle OPM)$
$={120}^{\circ }$