Question

In the given figure, A, B and C are three points on a circle with centre O such that $\angle BOC={30}^{\circ }$ and $\angle AOB={60}^{\circ }$. If D is a point on the circle other than the arc ABC, find $\angle ADC$.

• A. $(\frac{90}{2}{)}^{\circ }$
• B. ${45}^{\circ }$
• C. $(\frac{60}{2}{)}^{\circ }$
• D. ${30}^{\circ }$

Answer 1

Answer: (A), (B)

The correct options are
A $(\frac{90}{2}{)}^{\circ }$
B ${45}^{\circ }$

It can be observed that
$\angle AOC=\angle AOB+\angle BOC={60}^{\circ }+{30}^{\circ }=90$

We know that angle subtended by an arc at the center is double the angle subtended by it any point on the remaining part of the circle.

$\text{So},\angle ADC=\frac{1}{2}\angle AOC$
$=\frac{1}{2}\times {90}^{\circ }={45}^{\circ }$

$\Rightarrow \angle ADC={45}^{\circ }$.