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Question

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

O is the center of the circle and BCO=30o.

In given figure join OB and AC.

In BOC,

CO=BO [ Radius of a circle ]

OBC=OCB=30o [ Angles opposite to equal sides are equal ]

BOC+OBC+OCB=180o [ By angles sum property of a triangle ]

BOC+30o+30o=180o

BOC=180o60o

BOC=120o

We know that, in a circle, the angle subtended by an arc at the center is twice the angle subtended by it at the remaining part of the circle.
BOC=2BAC

BAC=120o2=60o

BAE=CAE=30o [ AE is an angle bisector of A ]

BAE=x=30o
In ABE,

30o+EBA+90=180o

120o+EBA=180o

EBA=60o

ABD+y=60o

In a circle, the angle subtended by an arc at the center is twice the angle subtended by it at the remaining part of the circle.

12×AOD+y=60o

90o2+y=60o

45o+y=60o

y=60o45o

y=15o


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