# In the given fig. chords AB, BC and CD are equal and O is the centre of the circle. If angle ABC =120 degree , find the measure of 1. ∠BAC 2.∠BEC 3.∠COD 4.∠BOD

1
by Angelkajal
is it rd sharma
no.....reliable
ok

2016-02-29T17:12:33+05:30
In Triangle ABC
AB = AC Therefore Triangle ABC is an isosceles traingle

So,
∠BAC + ∠BCA + ∠ABC = 180° [Angle sum property]
But it is an isosceles traingle so base angles are equal
Therefore,

∠BAC = ∠BCA
Therefore,
∠BAC + ∠BAC + 120 = 180
2∠BAC = 180 - 120
∠BAC = 60/2
BAC = 30°

Now,
∠ECB = 90° [Angle subtended by the diameter on the circumference of cirlce is complemantary]

∠ECA + ∠BCA = 90 [BAC = BCA = 30]
∠ECA + 30 = 90
∠ECA = 60°

Triangle COB is equilateral epecified below
So,
∠CBO = ∠CBE = 60°
﻿In Traingle ECB
∠ECB + ∠CBE + ∠BEC = 180° [Angle sum property]
﻿90 + 60 + ∠BEC= 180
∠CBE = 180 - 9 - 60
∠BEC = 30°

Traingle EOD,DOC and COB are equilateral traingle as OE = OD = OC = OB all are radii of the circle and ED = DC = CB so all the 3 sides of all the 3 triangles are equal so by SSS Test All the 3 triangles are congruent
So,
∠EOD = ∠DOC = ∠COB ﻿[By C.P.C.T.C]
∠EOD + ∠DOC + ∠COB = 180° ﻿[EOB is a straight line]
∠EOD + ∠EOD + ∠EOD = 180°
3∠EOD = 180
∠EOD = 60 = ∠COD = ∠BOD

Therefore,
∠COD = ∠BOD = 60°

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Millions thanks !!!
Millions Welcome !!!
Srry Where i have wrote BOD its actually BOC and BOD = 120