Question

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

• A. ${50}^{\circ }$
• B. $\frac{1}{2}\angle AOB$
• C. ${60}^{\circ }$
• D. ${70}^{\circ }$

Answer 1

Answer: (A), (B)

The correct options are
A ${50}^{\circ }$
B $\frac{1}{2}\angle AOB$

In $\Delta$ OAB,
OA = OB [radii]
$\angle OAB=\angle OA\Rightarrow \angle OBA={40}^{\circ }$
[angles opposite to equal sides are equal]
Also,
$\angle AOB+{40}^{\circ }+{40}^{\circ }={180}^{\circ }$
[angle sum property of a triangle]
$\therefore \angle AOB+{40}^{\circ }+{40}^{\circ }={180}^{\circ }$
$\Rightarrow \angle AOB={180}^{\circ }-{80}^{\circ }={100}^{\circ }$
We know that, in a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
$\therefore \angle AOB=2\angle ACB$
$\Rightarrow {100}^{\circ }=2\angle ACB$
$\therefore \angle ACB=\frac{{100}^{\circ }}{2}={50}^{\circ }$