SSC CHSL 16 Apr 2021 Shift 2 Paper

Given, Since, $\mathit{A}\mathit{B}=\mathit{A}\mathit{C}$,
$\therefore \angle \mathit{A}\mathit{B}\mathit{C}=\angle \mathit{A}\mathit{C}\mathit{B}=2\mathit{x}$ (let)
Since, $\mathit{B}\mathit{E}$ is an angle bisector.
$\therefore \angle \mathit{E}\mathit{B}\mathit{C}=\frac{2\mathit{x}}{2}={\mathit{x}}^{\circ }$
Since, $\mathit{C}\mathit{E}$ is an angle bisector.
$\therefore \angle \mathit{A}\mathit{C}\mathit{E}=\angle \mathit{E}\mathit{C}\mathit{D}=\mathit{y}$ (let)


$\mathit{x}+2\mathit{x}+\mathit{y}+35={180}^{\circ }$
$[\because$ angle sum of triangle ]
$3\mathit{x}+\mathit{y}={180}^{\circ }-{35}^{\circ }={145}^{\circ }$
Since, $\mathit{B}\mathit{D}$ is a straight line.
$2\mathit{x}+2\mathit{y}={180}^{\circ }$
$\mathit{x}+\mathit{y}={90}^{\circ }$
$\therefore \mathit{y}={90}^{\circ }-\mathit{x}$
Put this value in Eq. (i)
$3\mathit{x}+90-\mathit{x}={145}^{\circ }$
$\therefore 2\mathit{x}={145}^{\circ }-{90}^{\circ }={55}^{\circ }$
$\therefore \angle \mathit{A}\mathit{B}\mathit{C}={55}^{\circ }$