Question

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

• A. ${30}^o.$
• B. ${50}^o.$
• C. ${60}^o.$
• D. ${120}^o.$

Answer 1

Answer: (C)

${60}^o.$

Given- $BC$ is the diameter of a circle with centre $O$.

$\angle ABC\&\angle ADC$ are the angles, subtended by arc $AC$ to the circumference of the same segment of the circle. $\angle BAO={60}^o$.

To find out- $\angle ADC=?$

Solution- In $\Delta BAO$ we have $OA=OB$ (radii of the same circle).

$\therefore \angle ABO=\angle AOB$. i.e $4$ $\angle ABO+\angle AOB=2\angle ABO$.

So $2\angle ABO+\angle BAO={180}^o$ (angle sum property of triangles). $⟹2\angle ABO+{60}^o={180}^o⟹\angle ABO={60}^o.$

Again $\angle ABO=\angle ADC$, since the angles, subtended by an arc to the circumference of the same segment of the circle, are equal

$\therefore \angle ADC={60}^o$.

Ans- Option $C$