Step-by-step explanation:
Given: line DE || line FG & AN is transversal.
[tex]m\angle ABE = m\angle BCG \\ ... (corresponding\:\angle s) \\ \\
\therefore m\angle BCG = x\\ \\ x + (3x + 40) \degree \: = 180\degree \\ ..(linear \: pair \: \angle s) \\ \\ \therefore \: 4x = 180\degree - 40\degree \\ \\ \therefore \: 4x = 140\degree \\ \\ \therefore \: x = \frac{140\degree }{4} \\ \\ \huge \red{ \boxed{\therefore \: x = 35\degree }}[/tex]
