Question

In a parallelogram $ABCD,\angle DAB={40}^{\circ }$ find the other angles of the parallelogram.

Answer 1

ABCD is a parallelogram

Given: $\angle DAB=\angle BCD={40}^0$ and $AD\parallel BC$

We know that the sum of consecutive angle are ${180}^0$

$\angle CBA+\angle DAB={180}^0$

$\therefore \angle CBA=180-40={140}^0$

We know that in parallelogram opposite angles are equal.

Therefore, $\angle ADC=\angle CBA={140}^0$

So angles of parallelogram are ${40}^0,{140}^0,{40}^0,{140}^0$