Question

In figure, $O$ is center of circle, $BD=OD$ and $CD\perp AB$, then find $\angle CAB$

Answer 1

We know that $BD=OD$
$\because OD=OB$ (radius of circle)
$\therefore BD=OD=OB$
Thus, $OBD$ will be equilateral triangle.
Each angle of equilateral triangle is ${60}^0$
$\therefore \angle BOD=\angle ODB=\angle OBD={60}^0$
By arc $BD$, angle subtended at center $\angle BOD$ will be double the angle subtended at remaining part of circle i.e., $\angle DAB$
$\angle DAB=\frac{1}{2}\times {60}^0$
$={30}^0$
We know that $AO,\angle CAD$ is of
$\therefore \angle CAB=\angle DAB={30}^0$
Thus, $\angle CAB={30}^0$