Question

In the parallelogram $ABCD$ is $\angle A={65}^{\circ }$, find $\angle B,\angle C$ and $\angle D$

Answer 1

Let $ABCD$ be a parallelogram in which $\angle A={65}^{\circ }$.
Since $AD\parallel BC$ we can treat $AB$ as a transversal. So,
$\angle A+\angle B={180}^{\circ }$
${65}^{\circ }+\angle B={180}^{\circ }$
$\angle B={180}^{\circ }-{65}^{\circ }$
$\angle B={115}^{\circ }$
Since the opposite angles of a parallelogram are equal, we have
$\angle C=\angle A={65}^{\circ }$ and $\angle D=\angle B={115}^{\circ }$
Hence, $\angle B={115}^{\circ },\angle C={65}^{\circ }$ and $\angle D={115}^{\circ }$