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In the adjoining firgure, O is the centr...

In the adjoining firgure, O is the centre of the circle and AB is its diamter. If `AB||CD angle ABC=25^(@)`, then the value of `angle CED` is

A

`80^(@)`

B

`50^(@)`

C

`25^(@)`

D

`40^(@)`

Text Solution

Verified by Experts

The correct Answer is:
d
Let us join, B,D and A,D
`:. AB||CD :. angle ABD+ angle CDB=180^(@).....(1)`
Again, in the circle with centre at `O, angle CED and angle CBD` are two angels in circle produced by the are `overset(frown)(CD) :. angle CED= angle CBD......(1)`
Now, `angle ABC=25^(@),` (Given) `:. angle BCD=25^(@) [ :. AB ||CD`, alternate angle]

Again `angle ADC= angle ABC '[ :. ` angles in the same segment of circle]
`:. AB` is a diameter `:. angle AOB` is the central angle and angle is circle is `angle ADB` both produced by the are `overset(frown)(AEB)`.
`:.angle ADB=(1)/(2) angle AOB(1)/(2) xx180^(@)=90^(@) [ :. angle AOB` is a straight angle]
Then `angle BDC= angle ADC+ angle ADB= 25^(@)+90^(@)=115^(@)`
`:."in" Delta BCD, angle CBD=180^(@)-(angle BCD+ angle BDC)=180^(@)-(25^(@)+115^(@))`
`=180^(@)-140^(@)=40^(@)`
`:.` from (1) we get `angle CED= angle CBD= 40^(@)`
`:. (d) ` is correct
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