Question

$\square$ABCD is a kite in which diagonal AC and diagonal BD intersect at point O. $\angle OBC={20}^o$ and $\angle OCD={40}^o$

find $\angle$ADC

Answer 1

$\square ABCD$ is a kite where $AC$ and $BD$ are diagonals intersect at point $O$.

$\angle OBC={20}^o$ and $\angle OCD={40}^o$.

We know, in kite diagonals intersect at right angles.

$\therefore$ $\angle AOD=\angle DOC=\angle BOC=\angle AOB={90}^o$

In $△DOC,$

$\Rightarrow$ $\angle DOC+\angle OCD+\angle ODC={180}^o$

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

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

$\therefore$ $\angle ODC={50}^o$

$\Rightarrow$ $AD=CD$ [ Adjacent sides are equal in length ]

$\Rightarrow$ $\angle DAC=\angle DCA={40}^o$ [ Angles opposite to equal sides are equal ]

In $△AOD,$

$\Rightarrow$ $\angle DAO+\angle AOD+\angle ADO={180}^o$

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

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

$\Rightarrow$ $\angle ADC=\angle ADO+\angle ODC$

$={50}^o+{50}^o$

$={100}^o$

$\therefore$ $\angle ADC={100}^o$.