Question

In figure, if $\angle OAB={40}^{\circ }$, then $\angle ACB$ is equal to ___.

• A. ${50}^{\circ }$
• B. ${40}^{\circ }$
• C. ${60}^{\circ }$
• D. ${70}^{\circ }$

Answer 1

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

We have OA = OB = radius of the circle.
Since angles opposite to equal sides are equal, we have
$\angle OAB=\angle OBA.$
$\therefore \angle OBA={40}^{\circ }$
Now, using angle sum property in $△OAB,$ we have
$\angle OAB+\angle OBA+\angle AOB={180}^{\circ }.$
$⟹\angle AOB={180}^{\circ }-({40}^{\circ }+{40}^{\circ })={100}^{\circ }$
Since the angle subtended by an arc at the centre of a circle is double the angle subtended by it on any remaining part of the circle,
$\angle ACB=\frac{1}{2}\angle AOB={50}^{\circ }$