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

RS Aggarwal & V Aggarwal - Mathematics

In Fig. 16.135, *O* is the centre of the circle, *Bo* is the bisector of ∠*ABC*. Show that *AB=AC.*

In Fig. 16.121, it is given that *O* is the centre of the circle and ∠*AOC*=150°. Find ∠*ABC*.

RD Sharma - Mathematics