Question

In the figure, $segAB$ and $segAD$ are the chords of the circle. $C$ is a point on the tangent of the circle at point $A$. If $m\left(arcAPB\right)={80}^0$ and $\angle BAD={30}^0$,

The $m\left(arcBQD\right)$ is :

• A. ${60}^0$
• B. ${40}^0$
• C. ${30}^0$
• D. ${80}^0$

Answer 1

The correct option is A ${60}^0$

Given-

$O$ is the centre of a circle with a tangent $AC$ at $A$.

$AB$ and $AD$ are two chords.

$m\left(arcAPB\right)={60}^o$ and $\angle BAD={30}^o$

To find out- $m\left(arcBQD\right)=?$

Solution-

We join $BO$ and $BD$.

Now $m\left(arcBQD\right)$ means $\angle BOD$.

We know that the angle subtended by a chord of a circle at the centre is double the angle subtended by the same chord at the circumference of the circle.

$\therefore \angle BOD=2\times \angle BAD=2\times {30}^o={60}^o$