Question

$ABCD$ is rhombus such that $ACB={40}^0$, then find the $ADB$

Answer 1

Here, $ABCD$ is rhombus.

$AC$ and $BD$ are diagonals of rhombus. Diagonals of rhombus perpendicularly bisect each other at point $O.$

$\therefore$ $\angle BOC={90}^o.$

$\Rightarrow$ $\angle ACB={40}^o$ [ Given ]

i.e. $\angle OCB={40}^o$

In $△OCB,$

$\Rightarrow$ $\angle BOC+\angle OCB+\angle CBO={180}^o.$

$\Rightarrow$ ${90}^o+{40}^o+\angle CBO={180}^o.$

$\Rightarrow$ ${130}^o+\angle CBO={180}^o.$

$\Rightarrow$ $\angle CBO={50}^o$

i.e. $\angle CBD={50}^o$

$AD\parallel BC$ and $BD$ is traversal.

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

$\therefore$ $\angle ADB={50}^o.$