Question

In the given figure, A, B and C are three points on the circle with centre O such that $\angle BOC={30}^{\circ }$ and $\angle AOB={60}^{\circ }$. If D is a point on the circle that is not on the arc ABC, then find $\angle ABC.$

• A. ${135}^{\circ }$
• B. ${45}^{\circ }$
• C. ${120}^{\circ }$
• D. $(\frac{60}{2}{)}^{\circ }$

Answer 1

The correct option is A ${135}^{\circ }$


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

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

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

$\therefore \angle ADC={45}^{\circ }$

Consider the cyclic quadrilateral ABCD. Since opposite angles are supplementary, we have
$\angle ADC+\angle ABC={180}^{\circ }.$
$⟹\angle ABC={180}^{\circ }-\angle ADC={180}^{\circ }-{45}^{\circ }={135}^{\circ }$