Question

In the given figure, $AC$ is the bisector of $\angle A$. If $AB=AC,AD=CD$ and $\angle ABC={75}^o$, find the value of $x$ and $y$.

Answer 1

In $△ABC,$

$\Rightarrow$ $AB=AC$ [ Given ]

$\Rightarrow$ $\angle ABC=\angle ACB$ [ Base angles of equal sides are also equal. ]

$\Rightarrow$ $\angle ABC={75}^o$

$\therefore$ $\angle ACB={75}^o$

$\Rightarrow$ $\angle ABC+\angle ACB+\angle BAC={180}^o$

$\Rightarrow$ ${75}^o+{75}^o+x={180}^o$

$\Rightarrow$ ${150}^o+x={180}^o$

$\Rightarrow$ $x={30}^o.$

It is given that, $AC$ bisects $\angle A$

$\therefore$ $\angle BAC=\angle CAD$

$\therefore$ $\angle CAD={30}^o$

In $△ACD,$

$\Rightarrow$ $AD=CD$ [ Given ]

$\Rightarrow$ $\angle CAD=\angle ACD$ [ Base angles of equal sides are also equal. ]

$\therefore$ $\angle ACD={30}^o$

$\Rightarrow$ $\angle CAD+\angle ACD+\angle CDA={180}^o$

$\Rightarrow$ ${30}^o+{30}^o+y={180}^o$

$\Rightarrow$ ${60}^o+y={180}^o$

$\Rightarrow$ $y={120}^o$

$\therefore$ The value of $x$ and $y$ are ${30}^o$ and ${120}^o.$