Given,
\(angle ADC = frac{1}{2}angle AOC\)We know that the angle subtended by the chord on the major area is half of the subtended on center
Therefore,
\(angle ADC = frac{1}{2}times 130=65degree\)It is given that ADCB is a cyclic quadrilateral
Therefore, the sun of opposite angles is \(180degree\) \(angle ABC+angle ADC=180degree\) \(angle ABC+65=180degree\) \(angle ABC=115degree\)
