Question

In the given figure, $AOB$ is a diameter of a circle with center $O$.

If $\angle BOD={120}^o$, then m$\angle ACD$ is

• A. ${30}^o$
• B. ${40}^o$
• C. ${60}^o$
• D. ${90}^o$

Answer 1

The correct option is A ${30}^o$

Given-

$O$ is the centre of a circle of which $AOB$ is a diameter.

$CD$ is another chord, intersecting $\overline{AOB}$.

$\angle BOD={120}^o$

To find out- $\angle ACD=?$

Solution-

$\angle AOD={180}^o-\angle BOD={180}^o-{120}^o={60}^o$ ....(linear pair).

Now $\angle AOD$ is the angle subtended by the minor $arcAD$ to the centre and $\angle ACD$ is the angle subtended by the minor arc $AD$ to the circumference at $C$.

$\therefore \angle ACD=\frac{1}{2}\angle ACD=\frac{1}{2}\times {60}^o={30}^o$