In the given figure, O is the centre of a circle and BCO = 30°. Find x and y.

Given:


In ΔEOC


By angle sum property


EOC + OEC + OCE = 180°


EOC + 90° + 30° = 180°


EOC = 180° – 90° – 30° = 60°


EOC = 60°


Here,


EOD = EOC + COD = 90°


EOC + COD = 90°


60° + COD = 90°


COD = 90° – 60° = 30°


Now,


AOC = AOD + COD = 90° + 30° = 120°


We know that ,


COD = 2×CBD


COD = CBD


CBD = × 120° = 60°


Consider ΔABE


By angle sum property


AEB + ABE + BAE = 180°


90° + 60° + BAE = 180°


BAE = 180° – 90° – 60° = 30°


x = 30°


We know that ,


AOC = 2×ABC


AOC = ABC


ABC = × 30° = 15°


y = 15°



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