Question

Question 9
In the figure, BC is a diameter of the circle and $\angle BAO={60}^{\circ }$. Then, $\angle ADC$ is equal to

(A) ${30}^{\circ }$
(B) ${45}^{\circ }$
(C) ${60}^{\circ }$
(D) ${120}^{\circ }$

Answer 1

(C) ${60}^{\circ }$

Given, $\angle BAO={60}^{\circ }$
In $\Delta AOB$
OA = OB
$\Rightarrow \angle OBA=\angle BAO$ [angles opposite to equal sides are equal]
$\Rightarrow \angle BAO=\angle OBA={60}^{\circ }$

We know that, $\angle ABC=\angle ADC$ [ angles in the same segment AC are equal]
$\therefore \angle ADC={60}^{\circ }$