Question

In the adjoining figure line l $\parallel$ line m and PQ is the transversal. If $\angle$PEB = 70$°$. Find the measures of each of the remaining angles.

Answer 1

$$\begin{gathered} \mathrm{Given}:l\mathit{}\parallel m \\ \mathrm{and}\angle \mathrm{PEB}\hspace{0.17em}=70° \\ \mathrm{We}\mathrm{have}: \\ \angle \mathrm{PEB}\hspace{0.17em}=\angle \mathrm{EFD}=70°(\mathrm{Corresponding}\mathrm{angles}) \\ \angle \mathrm{EFD}=\angle \mathrm{CFQ}=70°(\mathrm{Vertically}\mathrm{opposite}\mathrm{angles}) \\ \angle \mathrm{AEF}=\angle \mathrm{PEB}\hspace{0.17em}=70°(\mathrm{Vertically}\mathrm{opposite}\mathrm{angles}) \\ \mathrm{Also},\angle \mathrm{PEA}+\angle \mathrm{PEB}\hspace{0.17em}=180°(\mathrm{Linear}\mathrm{pair}) \\ \Rightarrow \angle \mathrm{PEA}=180°-70°=110° \\ \angle \mathrm{BEF}=\angle \mathrm{PEA}=110°(\mathrm{Vertically}\mathrm{opposite}\mathrm{angles}) \\ \angle \mathrm{EFC}+\angle \mathrm{EFD}=180°(\mathrm{Linear}\mathrm{pair}) \\ \Rightarrow \angle \mathrm{EFC}+70°=180° \\ \Rightarrow \angle \mathrm{EFC}=110° \\ \angle \mathrm{DFQ}=\angle \mathrm{EFC}=110°(\mathrm{Vertically}\mathrm{opposite}\mathrm{angles}) \end{gathered}$$