In the given figure, `AB||CD` and `angleCDP=35^(@)`. PD is produced downwards to meet AB at E and `angleBEF=75^(@)`. If `DQ||EF`, find `angleAED, angleDEF` and `anglePDQ`.

In the given figure, `AB||CD` and `angleCDP=35^(@)`. PD is produced do

1.	In the given figure, `AB\|\|CD` and `angleCDP=35^(@)`. PD is produced downwards to meet AB at E and `angleBEF=75^(@)`. If `DQ\|\|EF`, find `angleAED, angleDEF` and `anglePDQ`.
Answer» `AB\|\|CD` and PDF is the transversal. `:. angleAED=angleCDP=35^(@)` [corresponding `angles`]. Now, AB is a straight line and ray PDE stands on it. `:. angleAED+angleDEF+angleBEF=180^(@)` `implies 35^(@)+angleDEF+75^(@)=180^(@)` `implies angleDEF=(180^(@)-110^(@))=70^(@)`. Now, `DQ\|\|EF` and PDE is the transversal. `:. anglePDQ=angleDEF=70^(@)` [corresponding `angles`]. Thus, `angleAED=35^(@), angleDEF=70^(@)` and `anglePDQ=70^(@)`.