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AOB is a diamter of a circle. When the ...

AOB is a diamter of a circle. When the chords AC and BD are extended they meet at the point E. if `angle COD=40^(@)`, then the value of `angle CED` is

A

`40^(@)`

B

`80^(@)`

C

`20^(@)`

D

`70^(@)`

Text Solution

Verified by Experts

The correct Answer is:
D

Let us join B,C and C,D AOB is a diameter of the circle with centre at O.
`:. angleACB` is an angle is semicircle.
`:. angle ACB=90^(@).....(1)`
Again `angle BCE=180^(@)-angle ACB=180^(@)-angle 90^(@)` [ from (1)]
`=90^(@).......(2)`
Now, the front central angle produced by the chord CD is `angle COD=40^(@)` and the fron angle is circle is `angle CBD`
by theorem -34 `angle CBD=(1)/(2) angle COD`
or, `angle CBE= (1)/(2) xx 40^(@)=20^(@)....(3) [ :. angle CBD and angle CBE` are same angles]
So from `Delta BCE` we get, `angle BEC+angle BCE+ angle CBE=180^(@)`
`or, angle BEC+90^(@)+20^(@)=180^(@)` [ from (2) and (3)]
or `angle BEC=180^(@)-110^(@)=70^(@)`
`:. angle BEC=70^(@) rArr (d)` is correct.
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