Question

In a trapezium ABCD, side AB is parallel to side DC. If $angleA={78}^{\circ }$ and $C={120}^{\circ }$, find angles B and D.

Answer 1

Given,
Trapezium ABCD,
$\angle A={78}^{\circ }$ and $\angle C={78}^{\circ }$
AB || CD.

Thus,
$\angle A+\angle D={180}^{\circ }$ ---(1) [Co-interior angles]
Similarily
$\angle B+\angle C={180}^{\circ }$ ---(2)

By using equation (1) and value of $\angle A$,
${78}^{\circ }+\angle D={180}^{\circ }$ $(\therefore \angle A={78}^{\circ })$
$\therefore \angle D={102}^{\circ }$

By using equation (2) and value of $\angle C$,
$\angle B+{120}^{\circ }={180}^{\circ }$ $(\because \angle C={120}^{\circ })$
$\therefore \angle B={60}^{\circ }$

Hence, value of $\angle B$ and $\angle D$ is ${60}^{\circ }$ and ${102}^{\circ }$ respectively.