Question

In the figure, if AB||HF and DE||FG, then the measure of $\angle$ FDE is

Answer 1

In the figure,

AB||HF, DE||FG
$\therefore HF||AB$
$\angle 1={28}^0$ (Corresponding angles)
But $\angle 1+\angle FDE+{72}^0={180}^0$
(Angles of a straight line)
$\Rightarrow {28}^0+\angle FDE+{72}^0={180}^0$
$\Rightarrow \angle FDE+{100}^0={180}^0$
$\Rightarrow \angle FDE={180}^0-{100}^0={80}^0$