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Question

In the given figure, O is the centre of the circle and BCO=30o. Find x and y.
1715491_0ff02c4cc82045508d1a90f9dd50db46.png

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Solution

From the figure we know that AOD and OEC from right angles

So we get

AOD=OEC=90o

We know that OD||BC and OC is a transversal

From the figure we know that AOD and OEC are corresponding angles

AOD=OEC

We know that DOC and OCE are alternate angles

DOC=OCE=30o

Angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference

So we get

DOC=2DBC

It can be written as

DBC=12DOC

By substituting the values

DBC=302

By division

y=DBC=15o

In the same way

ABD=12AOD

By substituting the values

ABD=902

By division

ABD=45o

We know that

ABE=ABC=ABD+DBC

So we get

ABE=ABC=45o+15o

By addition

ABE=ABC=60o

Consider ABE

Using the angle sum property

BAE+AEB+ABE=180o

By substituting the values

x+90o+60o=180o

On further calculation

x=180o90o60o

By subtraction

x=180o150o

So we get

x=30o

Therefore , the value of x is 30o and y is 15o.

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