Question

In the figure, $ABC$ is a triangle in which $AB=AC$. Find $x^o$ and $y^o$

Answer 1

Consider the figure,

$\angle ACB+\angle ACE={180}^o$ [Adjacent Angle]

$\Rightarrow x+x+48={180}^o$

$\therefore 2x=132$

$\therefore x={66}^o$

Given that $AB=AC$

$\Rightarrow \angle ABC=\angle ACB=x={66}^o$

Consider $△ABC$,

By Angle Sum Property, sum of all interior angles $={180}^o$

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

$\Rightarrow 66+66+\angle BAC={180}^o$

$\therefore \angle BAC={48}^o$

$\therefore \angle y=180-\angle BAC=180-48={132}^o$

$\therefore y={132}^o$