Question

In the figure, $AB\parallel CD$ and $\angle F={30}^0$ and find $\angle ECD$.

Answer 1

In $△FAE$

$\angle FAE+\angle AEF+\angle EFA={180}^{\circ }$ by angle sum property

$\Rightarrow {90}^{\circ }+\angle AEF+{30}^{\circ }={180}^{\circ }$

$\Rightarrow \angle AEF={180}^{\circ }-{120}^{\circ }={60}^{\circ }$

$\angle ECD=\angle AEF={60}^{\circ }$(vertically opposite angles)

$\angle ECD+\angle CEB={180}^{\circ }$(angles on the same side)

$\Rightarrow \angle ECD+{60}^{\circ }={180}^{\circ }$

$\therefore \angle ECD={180}^{\circ }-{60}^{\circ }={120}^{\circ }$