Question

In the adjoining figure 1.89, AB || CD. Find the value of x.

Answer 1

Given: $AB\parallel CD,\angle DCE=100°\mathrm{and}\angle ABE=120°$
Construction: Extend AB and CD.



We have,
$$\begin{gathered} \angle \mathrm{ABE}+\angle \mathrm{GBE}=180°(\mathrm{Linear}\mathrm{pair}) \\ \Rightarrow 120°+\angle \mathrm{GBE}=180° \\ \Rightarrow \angle \mathrm{GBE}=60°......\left(\mathrm{i}\right) \\ \mathrm{Also},\angle \mathrm{DCG}+\angle \mathrm{CGF}=180°(\mathrm{Angles}\mathrm{on}\mathrm{the}\mathrm{same}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{transversal}\mathrm{of}\mathrm{two}\mathrm{parallel}\mathrm{lines}\mathrm{are}\mathrm{supplementary}) \\ \Rightarrow 100°+\angle \mathrm{CGF}=180° \\ \Rightarrow \angle \mathrm{CGF}=80° \\ \mathrm{and}\angle \mathrm{BGE}=\angle \mathrm{CGF}=80°(\mathrm{Vertically}\mathrm{opposite}\mathrm{angles}).....\left(\mathrm{ii}\right) \\ \mathrm{In}△\mathrm{BGE},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{GBE}+\angle \mathrm{BGE}+x°=180°(\mathrm{Angle}\mathrm{sum}\mathrm{property}) \\ \Rightarrow 60°+80°+x°=180° \\ \therefore x°=40° \end{gathered}$$