Question

In the given figure, `AB||CD||EF`, `angleDBG`=x, `angleEDH`=y, `angleAEB`=z, `angleEAB`=`90^(@)` and `angleBEF=65^(@)`. Find the values of x, y, and z.

Answer 1

`EF||CD` and ED is the transversal.
`:. angleFED+angleEDH=180^(@)` [co-interior `angles`]
`implies 65^(@)+y=180^(@) implies y=(180^(@)-65^(@))=115^(@)`
Now, `CH||AG` and DB is the transversal.
`:. x=y=115^(@)` [corresponding `angles`].
Now, ABG is a straight line.
`:. angleABE+angleEBG=180^(@)` [linear pair]
`implies angleABE+x=180^(@)`
`implies angleABE+115^(@)=180^(@)`
`implies angleABE=(180^(@)-115^(@))=65^(@)`
We know that the sum of the angles of a triangle is `180^(@)`. From `Delta EAB`, we get
`angleEAB+angleABE+angleBEA=180^(@)`
`implies 90^(@)+65^(@)+z=180^(@)`
`implies z=(180^(@)-155^(@))=25^(@)`.
`:. X=115^(@), y=115^(@) and z=25^(@)`.