In the given figure, O is the centre of a circle in which AOC = 100°. Side AB of quad. OABC has been produced to D. Then, CBD = ?

Given: AOC = 100°


Here,


(Exterior AOC) = 360° – (interior AOC)


(Exterior AOC) = 360° – 100°


(Exterior AOC) = 260°


We know that,


(Exterior AOC) = 2 × ADC


260° = 2 × ABC


ABC = = 130°


ABC = 130°


Here,


ABD = ABC + CBD


180° = 130° + CBD


CBD = 180° – 130° = 50°


CBD = 50°

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