Question

In the given figure, AB || PQ. Find the values of x and y.

Answer 1

Given, $AB\parallel PQ$.
Let CD be the transversal cutting AB and PQ at E and F, respectively.
Then,
$$\begin{gathered} \angle CEB+\angle BEG+\angle GEF=180°\left[\text{Since CD is a straight line}\right] \\ \Rightarrow 75°+20°+\angle GEF=180° \\ \Rightarrow \angle GEF=85° \end{gathered}$$
We know that the sum of angles of a triangle is $180°$.
$$\begin{gathered} \therefore \angle GEF+\angle EGF+\angle EFG=180 \\ \Rightarrow 85°+x+25°=180° \\ \Rightarrow 110°+x=180° \\ \Rightarrow x=70° \end{gathered}$$
And
$$\begin{gathered} \angle FEG+\angle BEG=\angle DFQ\left[\text{Corresponding Angles}\right] \\ \Rightarrow 85°+20°=\angle DFQ \\ \Rightarrow \angle DFQ=105° \\ \angle EFG+\angle GFQ+\angle DFQ=180°\left[\text{Since CD is a straight line}\right] \\ \Rightarrow 25°+y+105°=180° \\ \Rightarrow y=50° \\ \therefore x=70°\text{and}y=50° \end{gathered}$$