In the given figure, O is the centre of the circle and `/_ BCO = 30^(@) `. Find x and y .

Answer

`x= 30^(@), y= 15^(@)`

Explanation

`/_ AOD = /_ OEC = 90^(@) ` ( given )
But, these are corresponding `/_ s`
`:. OD || BC `.
Now, `OD || BC ` and OC is the transversal .
`:. /_ DOC = /_ BCO = 30^(@)`
`:. /_ DBC = (1)/(2) /_ DOC = 15^(@) implies y = 15^(@)`
`/_ AOD = 90^(@) ` ( given )
Now, `/_ ABD = (1)/(2) /_ AOD = ((1)/(2) xx 90^(@)) = 45^(@) `
In `Delta ABE `, we have
`/_ ABE = ( /_ ABD + /_ DBC ) = ( 45^(@) + 15^(@)) = 60^(@) , /_ AEB = 90^(@)`
`:. /_ BAE = 180^(@) - ( 90 ^(@) + 60^(@)) = 30^(@) implies x = 30^(@)`