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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