Question

In the given fig. $ABCD$ is a parallelogram. Find the angles $x$ and $y$.

Answer 1

Here, $ABCD$ is a parallelogram.

$\angle CBD={40}^o$ and $\angle BAD={80}^o.$

$AD\parallel BC$ and $BD$ is a transversal.

$\therefore$ $\angle ADB=\angle CBD$ [ Alternate angles ]

$\therefore$ $\angle ADB={40}^o.$

$\therefore$ $y={40}^o.$

In $△ADB,$

$\Rightarrow$ $\angle ADB+\angle DBA+\angle BAD={180}^o$

$\Rightarrow$ ${40}^o+x+{80}^o={180}^o$

$\Rightarrow$ ${120}^o+x={180}^o$

$\Rightarrow$ $x={180}^o-{120}^o$

$\Rightarrow$ $x={60}^o$

We get, $x={60}^o$ and $y={40}^o.$