Question

In the given figure, $\angle AOB={90}^{\circ }and\angle ABC={30}^{\circ }.Then,\angle CAO=?$

(a) ${30}^{\circ }$

(b) ${45}^{\circ }$

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

(d) ${90}^{\circ }$

Answer 1

(c) 60°
We have:

In triangle $△AOB$

$OA=OB$

$\therefore \angle OAB+\angle OBA+\angle AOB={180}^0\Rightarrow 2\angle OAB=180-90=90\Rightarrow OAB=\frac{9}{2}={45}^0$

We know that the angle subtended at the centre of the circle by an arch is twice the angle subtended at the circumference by the same arc.

$\therefore \angle ACB=\frac{1}{2}\angle AOB=\frac{90}{2}={45}^0$

Again in $△ABC$

$\angle ACB+\angle CBA+\angle BAC={180}^0\Rightarrow 45+30+\angle BAC={180}^0\Rightarrow \angle BAC=180-75={105}^0\therefore \angle CAO=\angle BAC-\angle OAB\Rightarrow CAO={105}^0-{45}^0={60}^0$