\\\\(\\angle ADC = \\dfrac{1}{2} \\angle AOC\\\\)(Angle subtended by a chord on mayor area is half of that subtended on center)\\\\(\ herefore \\angle ADC = \\dfrac{1}{2} \ imes 130 = 65^{\\circ}\\\\)Now, ADCB is cyclic quadrilateral \\\\(\\Rightarrow \\\\) Sum of opposite Angles is \\\\(180^{\\circ}\\\\)\\\\(\ herefore \\angle ABC + \\angle ADC = 180^{\\circ}\\\\)\\\\(\\angle ABC + 65= 180^{\\circ}\\\\)\\\\(\\angle ABC = 115^{\\circ}\\\\)