Question :
In the adjoining figure $\angle AOC=140^{\circ}$, where O is the centre of the circle then $\angle ABC$ is equal to: 
Option 1: $110^{\circ}$
Option 2: $100^{\circ}$
Option 3: $90^{\circ}$
Option 4: $40^{\circ}$
Correct Answer: $110^{\circ}$
Solution :
Given: $\angle AOC = 140^{\circ}$
Construction: Take D on the remaining arc.
We know that,
The angle formed by the chord at the centre is twice the angle formed in the same arc segment.
$\angle AOC = 2\angle ADC$
⇒ $140^{\circ} = 2\angle ADC$
$\therefore \angle ADC = 70^{\circ}$
Now, the sum of the opposite angle of the cyclic quadrilateral is $180^{\circ}$.
$\angle ABC+\angle ADC=180^{\circ}$
⇒ $\angle ABC+70^\circ=180^{\circ}$
$\therefore\angle ABC=110^{\circ}$
Hence, the correct answer is $110^{\circ}$.
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