Question

In the given figure, AB || CD || EF, EA ⊥ AB and BDE is the transversal, such that ∠DEF = 55°. Then, ∠AEB = ?
(a) 35°
(b) 45°
(c) 25°
(d) 55°

Answer 1

(a) 35°

We have :
$EA\perp AB$
Therefore, $\angle BAE=90°$
And AB || EF and BE is transversal
$$\begin{gathered} \angle BAE+\angle AEF=180° \\ \Rightarrow 90°+\angle AEF=180° \\ \therefore \angle AEF=90° \end{gathered}$$

$$\begin{gathered} \Rightarrow \angle AEB+\angle BEF=90° \\ \Rightarrow \angle AEB+\angle DEF=90° \\ \Rightarrow \angle AEB+55°=90° \\ \Rightarrow \angle AEB=35° \end{gathered}$$