Question

Question 10
In the figure, if $\angle AOB={90}^{\circ }and\angle ABC={30}^{\circ },then\angle CAO$ is equal to

(A) ${30}^{\circ }$
(B) ${45}^{\circ }$
(C) ${90}^{\circ }$
(D) ${60}^{\circ }$

Answer 1

(D) ${60}^{\circ }$

$Given,\angle AOB={90}^{\circ }and\angle ABC={30}^{\circ }$
We know, that, in a circle the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle
$\therefore \angle AOB=2\angle ACB$
$\Rightarrow {90}^{\circ }=2\angle ACB[\therefore \angle AOB={90}^{\circ },given]$
$\Rightarrow \angle ACB={45}^{\circ }$
$Also,AO=OB$
$\angle ABO=\angle BAO$ [angle opposite to equal sides are equal] .....(i)

In $\Delta OAB,\angle OAB+\angle ABO+\angle BOA={180}^{\circ }$ [angle sum property of a triangle]
$\angle OAB+\angle OAB+{90}^{\circ }={180}^{\circ }$ [from Eq. (i)]
$\Rightarrow 2\angle OAB={180}^{\circ }-{90}^{\circ }$
$\Rightarrow \angle OAB=\frac{{90}^{\circ }}{2}={45}^{\circ }$

In $\Delta ACB,\angle ACB+\angle CBA+\angle CAB={180}^{\circ }$ [ angle sum property of triangle]
$\therefore {45}^{\circ }+{30}^{\circ }+\angle CAB={180}^{\circ }$
$\Rightarrow \angle CAB={180}^{\circ }-{75}^{\circ }={105}^{\circ }$
$\therefore \angle CAO+\angle OAB={105}^{\circ }$
$\Rightarrow \angle CAO+\angle OAB={105}^{\circ }$
$\therefore \angle CAO={105}^{\circ }-{45}^{\circ }={60}^{\circ }$