Question

In the above figure, $O$ is the centre of the circle, $BD=OD$ and $CD$ $\perp$ $AB$.

Find $\angle CAB$.

• A. ${60}^o.$
• B. ${45}^o.$
• C. ${30}^o.$
• D. ${15}^o.$

Answer 1

The correct option is C ${30}^o.$

Given- $BD$ is a chord of the circle with centre $O$, is equal to the radius $OD$ of the circle.

The diameter $AOB\perp DC.$

$AC$ and $BC$ are two other chords.

To find out-

$\angle CAB=?$

Solution:

In $△ODB$, we have

$OD=OB$ .....(radii of the same circle) $=BD$.

$\therefore$ $\Delta ODB$ is equilateral.

i.e all the angles $={60}^o$

$\therefore \angle OBD={60}^o$

$\therefore \angle ABD+\angle BDC+{90}^o={180}^o$ .....(angle sumd property of triangles).

$\Rightarrow \angle BDC={180}^o-{90}^o-{60}^o={30}^o$

Now the chord $BC$ subtends $\angle BAC$and $\angle BDC$ at the cicumference.

$\therefore \angle BAC=\angle BDC={30}^o$

The angles subtended by a chord of a circle at the circumference are equal.