AOCB is a quadrilateral in circle with centre O.

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 \(180degree\) \(angle ABC+angle ADC=180degree\) \(angle ABC+65=180degree\) \(angle ABC=115degree\)

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