Question

In Fig. 3, if $\angle OAB={40}^{\circ },then\angle ACB$ is equal to :

• A. ${50}^{\circ }$
• B. ${40}^{\circ }$
• C. ${60}^{\circ }$
• D. ${57}^{\circ }$

Answer 1

The correct option is A ${50}^{\circ }$

$OA=OB$ (Radii of same circle)

$\angle OBA=\angle OAB={40}^{\circ }$

In Triangle OAB by Angle Sum Property of Triangle ;

$\angle AOB+\angle OBA+\angle OAB={180}^{\circ }$

$\angle AOB={180}^{\circ }-{40}^{\circ }-{40}^{\circ }={100}^{\circ }$

We know $2.\angle ACB=\angle AOB$

therefore

$\angle ACB=\frac{{100}^{\circ }}{2}$

$\angle ACB={50}^{\circ }$

So, option A is the answer.