Question

In the adjoining figure $AOB$ is a diameter $MPQ$ is a tangent at $P$ then the value of $\angle MPA$ is equal to

• A. ${25}^{\circ }$
• B. ${26}^{\circ }$
• C. ${27}^{\circ }$
• D. ${30}^{\circ }$

Answer 1

The correct option is D ${30}^{\circ }$

From the above figure , it is clear that $\Delta OAP$ is equilateral triangle .

In $\Delta AOP$ , $\angle AOP={60}^o$, also $\angle OAP=\angle OPA={60}^O$ ....($AO=PO=r$, hence opposite angle are same )

Now it is clear that $\angle MPO={90}^o$ ,

Hence $\angle MPA={90}^o-{60}^o={30}^o$