Question

Consider the figure where $\angle AOB={90}^{\circ }$ and $\angle ABC={30}^{\circ }.$ Then, $\angle CAO$ = .

• A. 50°
• B. 65°
• C. 60°
• D. 40°

Answer 1

The correct option is C 60°

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

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

Using angle sum property in $\Delta ACB,$ we have
$\angle ACB+\angle CBA+\angle CAB={180}^{\circ }.$
$\therefore {45}^{\circ }+{30}^{\circ }+\angle CAB={180}^{\circ }$
$\Rightarrow \angle CAB={180}^{\circ }-{75}^{\circ }={105}^{\circ }$
But, $\angle CAO=\angle CAB-\angle OAB$
$={105}^{\circ }-{45}^{\circ }={60}^{\circ }.$
i.e., $\angle CAO={60}^{\circ }$