In the given figure, `AB` is a diameter of the circle with centre `O`, `AC` and `BD` produced meet at `E` and `/_ COD = 40^(@)` . Calculate `/_ CED`.

Explanation

Since `ACE` is a straight line, we have
`/_ ACB + /_BCE = 180^(@)`
`implies 90^(@) + /_ BCE = 180^(@)`
`[ :' /_ ACB` is in a semicircle ]
`implies /_BCE = 90^(@)`
Alao, `/_ DBC = (1)/(2) /_ COD = ((1)/(2) xx 40^(@)) = 20^(@)`
[ angle at centre ` = 2 xx ` angle at a point on a circle ]
`implies /_ EBC + /_ BCE + /_ CEB = 180^(@)`
`implies 20^(@) + 90^(@) + /_ CED = 180^(@) ` `[:' /_ CEB = /_ CED ]`
`implies /_ CED =180^(@) - 110^(@) = 70^(@)`
Hence, `/_ CED = 70^(@)`