Question

In the figure , $ABCD$ is a parallelogram in which $\angle DAB={75}^{\circ }and\angle DBC={60}^{\circ }$. Compute $\angle CDBand\angle ADB.$

Answer 1

In a parallelogram $ABCD$, the sum of consecutive angles is equals to ${180}^{\circ }$.
So, $\angle A+\angle B={180}^{\circ }$
But, $\angle A={75}^{\circ }$

$\therefore \angle B={180}^{\circ }-\angle A$
$={180}^{\circ }-{75}^{\circ }={105}^{\circ }$
But $\angle DBC={60}^{\circ }$
$\therefore DBA={105}^{\circ }-{60}^{\circ }={45}^{\circ }$

Since, $AB||CD$ and $BD$ acts as a transversal.
From the alternate interior angles property,
$\angle CDB=\angle DBA$
= ${45}^{\circ }$
Similar way for other set of alternate pair of angles,
we get $\angle ADB=\angle DBC={60}^{\circ }$