# 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,

$angle ADC = frac{1}{2}times 130=65degree$

Therefore, the sun of opposite angles is $180degree$ $angle ABC+angle ADC=180degree$ $angle ABC+65=180degree$ $angle ABC=115degree$

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