Question

In the given figure, `AB||CD, angleABO=40^(@), angleCDO=35^(@)`. Find the value of the reflex `angleBOD` and hence the value of x.

Answer 1

Through O, draw `EO||AB||Cd`.
Then, `angleEOB+angleEOD=x^(@)`.
Now, `AB||EO` and BO is the transversal.
`:. angleABO+angleBOE=180^(@)` [consecutive int. `angles`]
`implies 40^(@)+angleBOE=180^(@)`
`implies angleBOE=(180^(@)-40^(@))=140^(@)`.
Again, `CD||EO` and OD is the transversal.
`:. angleEOD+angleODC=180^(@)`
`implies angleEOD+35^(@)=180^(@)`
`implies angleEOD=(180^(@)-35^(@))=145^(@)`.
`:.` reflex `angleBOD=x^(@)=(angleBOE+angleEOD)`
`=(140^(@)+145^(@))=285^(@)`.
Hence, `x=285`.