The opposite sides of a quadrilateral are parallel. If one angle of the quadrilateral is $60^o$, find the other angles.

Given:

The opposite sides of a quadrilateral are parallel and one angle of the quadrilateral is $60^o$.

To do:

We have to find the other angles.

Solution:

Let $AB \parallel DC$ and $AD \parallel BC$ and $\angle A = 60^o$ in quadrilateral $ABCD$.

$AB \parallel DC$ and $AD \parallel BC$

This implies,

$ABCD$ is a parallelogram.

Therefore,

$\angle A + \angle B = 180^o$                      (Co-interior angles are supplementary)

$60^o + \angle B = 180^o$

$\angle B = 180^o-60^o= 120^o$

Opposite angles of a parallelogram are equal.

Therefore,

$\angle C = \angle A = 60^o$

$\angle D = \angle B = 120^o$

Hence, the other angles are $120^o, 60^o$ and $120^o$.