A quadrilateral ABCD is a trapezium in which side AB is parallel to side DC. If $\angle A = \angle B = {40^ \circ }$, what are the measures of the other two angles?

Hint: In trapezium, the sum of two adjacent angles made by parallel lines is ${180^ \circ }$. Use this theorem for the pair of angles made by parallel sides and find the required angles.

Complete step by step answer:

From the above figure, ABCD is a trapezium in which AB is parallel to DC and we have:
$ \Rightarrow \angle A = \angle B = {40^ \circ }.$
We know that, in trapezium, the sum of two adjacent angles made by parallel lines is ${180^ \circ }$. Thus, according to this rule, we have:
$ \Rightarrow \angle A + \angle D = {180^ \circ }$ (sum of interior angle on the same side of the transversal is ${180^ \circ }$)
$
   \Rightarrow \angle D = {180^ \circ } - \angle A \\
   \Rightarrow \angle D = {180^ \circ } - {40^ \circ } \\
   \Rightarrow \angle D = {140^ \circ } \\
$
Similarly for other two angles:
$
  \angle B + \angle C = {180^ \circ } \\
   \Rightarrow {40^ \circ } + \angle C = {180^ \circ } \\
   \Rightarrow \angle C = {140^ \circ } \\
$

Hence the measure of the other two angles is ${140^ \circ }$ each.

Note: If in a trapezium, non-parallel sides are equal in length then in that case, the opposite angles of the trapezium will be supplementary to each other. And such trapezium is called isosceles trapezium.