Question

In the given figure, $O$ is the centre of the circle and $\angle ABC=60^{\circ }$ then the measure of $\angle CDB$ (in degrees) is ___

Answer 1

Answer: 30

' $O^{\prime }$ is the centre of the circle with diameter $AB$.
$\angle ACB=90^{\circ }$
$\{\because$ Angle subtended on the area of the semicircle is $90^{\circ }\}$
Given: $\angle ABC=60^{\circ }$
Now, in $△ACB$
$\angle ACB+\angle CAB+\angle CBA=180^{\circ }$
$\Rightarrow 90^{\circ }+\angle CAB+60^{\circ }=180^{\circ }$
$\Rightarrow \angle CAB=180^{\circ }-150^{\circ }=30^{\circ }$
$\angle CAB=\angle CDB=30^{\circ }$
$\{\because$ Angles in the same segment of a circle are equal.$\}$