Question

In the adjoining figure, ABCD is a parallelogram in which $\angle DAB={80}^{\circ }and\angle DBC={60}^{\circ }$. Calculate $\angle CDBand\angle ADB$.

Answer 1

Given:

$ABCD$ is parallelogram and $\angle DAB={80}^{\circ }$, $\angle DBC={60}^{\circ }$
To find: Measure of $\angle CDB$ and $\angle ADB$

In parallelogram $ABCD$, $AD||BC$
$\Rightarrow \angle DBC=\angle ADB={60}^{\circ }$...(i) (Alternate interior angles)
As $\angle DAB$ and $\angle ADC$ are adajacent angles,

$\angle DAB+\angle ADC={180}^{\circ }$
$\therefore \angle ADC={180}^{\circ }-\angle DAB$
$\Rightarrow ADC={180}^{\circ }-{80}^{\circ }={100}^{\circ }$
Also, $\angle ADC=\angle ADB+\angle CDB$
$\therefore \angle ADC={100}^{\circ }$
$\Rightarrow \angle ADB+\angle CDB={100}^{\circ }$...(ii)
From (i) and (ii), we get:
${60}^{\circ }+\angle CDB={100}^{\circ }$

$\Rightarrow \angle CDB={100}^{\circ }-{60}^{\circ }$

$\Rightarrow \angle CDB={40}^{\circ }$
Hence, $\angle CDB={40}^{\circ }$ and $\angle ADB={60}^{\circ }$.