Lines EOF and AOB intersect each other at the point O.
`:. angleBOE=angleAOF=5y` ...(i)
[vertically opposite angles]
Also, the sum of all the angles formed on the upper side of COD at the point O is `180^(@)`.
`:. angleDOB+angleBOE+angleEOC=180^(@)`
`implies 2y+5y+2y=180^(@)` [using (i)]
`implies 9y=180^(@) implies y=20^(@)`.
Hence, `y=20^(@)`.