Question

AB$||$CD, FE$\perp$EC, AFE$={55}^o$ find $\angle a$.

Answer 1

Draw a line FL passing through E, parallel to B

$\Rightarrow FL||AB||CD$

Now $\angle AFE=\angle FEL$ (Alternate interior angles)

$\Rightarrow \angle FEL={55}^o$ Since $AB||FL$

Now $\angle FEL+\angle LEC=\angle FEC$

$\angle FEC+\angle LEC={90}^o$ ($\angle FEC={90}^o$(given))

$\Rightarrow \angle LEC={90}^o-{55}^o$

$\angle LEC={35}^o$

$\angle LEC=\angle ECP$ [Alternate interior angles]

$\angle ECP={35}^o$

$\Rightarrow a=\angle ECP+\angle PCD$

$a={35}^o+{180}^o$ ($[\angle PCD={180}^o]$ $\angle PCD$ angle is straight)

$a={215}^o$