Q. 194.0( 24 Votes )
In the given figure, O is the centre of a circle and ∠BCO = 30°. Find x and y.
Answer :
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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