Question

In given 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

The correct option is C ${60}^{\circ }$

In $△AOB$,

$OA=OB$

$=>\angle OAB=\angle OBA={60}^0$

We know that the exterior angle of a triangle is equal to sum of the two opposite angles of the triangle.

$\angle AOC=\angle OAB+\angle OBA$

$\angle AOC={60}^0+{60}^0$

$\angle AOC={120}^0$

We know that the angle subtended at the centre of the circle by an arch is twice the angle subtended at the circumference by the same arc.

$\therefore \angle ADC=\frac{1}{2}\angle AOC$

$=\frac{120}{2}$

$={60}^0$