In the given figure, `AB` is a diameter of a circle with centre `O`. If `ADE` and `CBE` are straight lines, meeting at `E` such that `/_ BAD = 35^(@)` and `/_ BED = 25^(@)` , find (i) `/_DBC , (ii) /_ BCD` and (ii) `/_ CDB`.

Explanation

Join `AC` and `BD`.
Now, `/_ADB = 90^(@)` ( angle in a semicircle ) .
Since `ADE` is a straight line, we have
`/_ ADB + /_ EDB = 180^(@) implies 90^(@) + /_ EDB = 180^(@) implies /_ EDB = 180^(@) - 90^(@) = 90^(@)`
In `Delta DBE`, we have
`/_ DBE + /_ BED + /_EDB = 180^(@) `
`implies /_DBE +25^(@) + 90^(@) = 180^(@)`
`implies /_DBE = 180^(@) -115^(@) = 65^(@)`.
(i) Since `CBE` is a straight line, we have
`/_ DBE+ /_DBE= 180^(@) implies /_DBC +65^(@) = 180^(@) implies /_DBC = 180^(@)-65^(@) = 115^(@)`
(ii) `/_ BCD = /_ BAD = 35^(@) ` [ angles in the same segment] .
(iii) In `Delta DBC` , we have
`/_ DBC + /_ BCD + /_ CDB = 180^(@)`
`implies 115^(@) + 35^(@) + /_ CDB = 180^(@)`
`implies /_CDB = (180^(@) -150^(@)) = 30^(@)`