In the given figure, O is the centre of a circle. IF `/_ AOD = 140^(@)` and `/_ CAB = 50^(@)`, calculate.
`(I) /_ EDB , (ii) /_ FBD`.

Answer

`(i) /_ EDB = 50^(@) , (ii) /_ EBD = 110^(@)`

Explanation

`/_ BOD = 180^(@) - /_ AOD = 180^(@) - 140^(@) = 40^(@)`
`OB= OD implies /_ OBD = /_ ODB = 70^(@)`
`/_ CAB + /_ BDC = 180^(@) [ :' ` ABDC is a cyclic ]
`implies /_ CAB + /_ ODB + /_ ODC = 180^(@)`
`implies 50^(@) + 70^(@) + /_ ODC = 180^(@)`
`implies /_ ODC = 60^(@)`
`:. /_ EDB = 180^(@) - (60^(@) + 70^(@)) = 50^(@)`
and `/_ EBD =180^(@) - /_ OBD = 180^(@) -70^(@) = 110^(@)`