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 $\angle$ABC

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 $△BOC,$

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

$\Rightarrow$ ${90}^o+{20}^o+\angle OCB={180}^o$

$\Rightarrow$ ${110}^o+\angle OCB={180}^o$

$\therefore$ $\angle OCB={70}^o$

$\Rightarrow$ $AB=BC$ [ Adjacent sides are equal in length ]

$\Rightarrow$ $\angle ACB=\angle BAC={70}^o$ [ Angles opposite to equal sides are equal ]

In $△AOB,$

$\Rightarrow$ $\angle ABO+\angle AOB+\angle OAB={180}^o$

$\Rightarrow$ $\angle ABO+{90}^o+{70}^o={180}^o$

$\Rightarrow$ $\angle ABO={20}^o$

$\Rightarrow$ $\angle ABC=\angle ABO+\angle OBC$

$={20}^o+{20}^o$

$={40}^o$

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