Question

Question 6
In the figure, if $\angle OAB={40}^{\circ }$, the $\angle ACB$ is equal to


(A) ${50}^{\circ }$
(B) ${40}^{\circ }$
(C) ${60}^{\circ }$
(D) ${70}^{\circ }$

Answer 1

(A) ${50}^{\circ }$

In $\Delta$ OAB,
OA = OB [both are the radius of a circle]
$\angle OAB=\angle OBA\Rightarrow \angle OBA={40}^{\circ }$ [angles opposite to equal sides are equal]
Also, $\angle AOB+\angle OBA+\angle OAB={180}^{\circ }$ [by angles sum property of a triangle]
$\angle AOB+{40}^{\circ }+{40}^{\circ }={180}^{\circ }$
$\therefore \angle AOB+{40}^{\circ }+{40}^{\circ }={180}^{\circ }$
$\Rightarrow \angle AOB={180}^{\circ }-{80}^{\circ }={100}^{\circ }$

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.
$\therefore \angle AOB=2\angle ACB$
$\Rightarrow {100}^{\circ }=2\angle ACB$
$\therefore \angle ACB=\frac{{100}^{\circ }}{2}={50}^{\circ }$