Question

In the adjoining figure, $AB\parallel CD$ and $l$ is transversal. Find the value of $x$ and $y$.

Answer 1

$\begin{array}{c}\angle EFD={60}^{\circ }=\angle LEB\left(correspondingangle\right)\\ \because AB\parallel CD\left[onstraightlineangleis{180}^{\circ }\right]\\ \Rightarrow \therefore \angle LEB+\angle BEP+\angle PEF={180}^{\circ }\\ \Rightarrow {60}^{\circ }+{40}^{\circ }+\angle PEF={180}^{\circ }\\ \Rightarrow \angle PEF={180}^{\circ }-100={80}^{\circ }\left[\because \angle EFD=\angle EFP+\angle PFD\right]\\ \Rightarrow y+25={60}^{\circ }\left[\because sumofallanglesoftriangleis{180}^{\circ }\right]\\ \Rightarrow y=60-25\\ y={35}^{\circ }Ans.\\ Thenthe\Delta FEP\\ \Rightarrow {80}^{\circ }+x+{25}^{\circ }={180}^{\circ }\\ \Rightarrow x+105={180}^{\circ }\\ \Rightarrow x={180}^{\circ }-105\\ x={75}^{\circ }Ans.\end{array}$