Question

In given fig. O is centre of the circle if $\angle AOC={130}^o$ then find $\angle ABC$.

Answer 1

$\angle ADC=\frac{1}{2}\angle AOC$

(Angle subtended by a chord on mayor area is half of that subtended on center)

$\therefore \angle ADC=\frac{1}{2}\times 130={65}^{\circ }$

Now, ADCB is cyclic quadrilateral $\Rightarrow$

Sum of opposite Angles is ${180}^{\circ }$

$\therefore \angle ABC+\angle ADC={180}^{\circ }$

$\angle ABC+65={180}^{\circ }$

$\angle ABC={115}^{\circ }$