Question

In Fig 10.9 $\angle AOB={90}^{\circ }and\angle ABC={30}^{\circ }then\angle CAO$ is equal to

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${90}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is D ${60}^{\circ }$

In $△AOB$

$OA=OB$

$\therefore \angle OAB+\angle OBA+\angle AOB={180}^0$

$=>2\angle OAB+{90}^0={180}^0$

$=>2\angle OAB={180}^0-{90}^0$

$={90}^0$

$=>\angle OAB=\frac{90}{2}$

$={45}^0$

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 $△ABC$

$\angle ACB+\angle CBA+\angle BAC={180}^0$

$=>{45}^0+{30}^0+\angle BAC={180}^0$

$=>\angle BAC={180}^0-{75}^0$

$={105}^0$

$\therefore \angle CAO=\angle BAC-\angle OAB$

$=>\angle CAO={105}^0-{45}^0$

$={60}^0$