SSC CGL 18 Aug 2021 Shift 3 Paper

Given, $\angle \mathit{A}\mathit{B}\mathit{C}$ and $\angle \mathit{C}\mathit{B}\mathit{D}$ are linear pairs.$\angle \mathit{A}\mathit{B}\mathit{C}+\angle \mathit{C}\mathit{B}\mathit{D}={180}^{\circ }$
$\mathit{B}\mathit{O}$ is bisector of $\angle \mathit{C}\mathit{B}\mathit{D}$
$\Rightarrow \angle \mathit{C}\mathit{B}\mathit{D}=2\angle 1$
$\Rightarrow \angle \mathit{A}\mathit{B}\mathit{C}+\angle \mathit{C}\mathit{B}\mathit{D}={180}^{\circ }$
$\Rightarrow 2\angle 1={180}^{\circ }-\angle \mathit{A}\mathit{B}\mathit{C}$
$\Rightarrow \angle 1={90}^{\circ }-\frac{1}{2}\angle \mathit{A}\mathit{B}\mathit{C}$ . . . (i)
Similarly, in $\angle \mathit{A}\mathit{C}\mathit{B}$ and $\angle \mathit{B}\mathit{C}\mathit{E}$
$\Rightarrow \angle \mathit{A}\mathit{C}\mathit{B}+\angle \mathit{B}\mathit{C}\mathit{E}={180}^{\circ }$
$\mathit{C}\mathit{O}$ is bisector of $\angle \mathit{B}\mathit{C}\mathit{E}$
$\Rightarrow \angle \mathit{A}\mathit{C}\mathit{B}+2\angle \mathit{Z}={180}^{\circ }$
$\Rightarrow 2\angle 2={180}^{\circ }-\angle \mathit{A}\mathit{C}\mathit{B}$
$\Rightarrow \angle 2={90}^{\circ }-\frac{1}{2}\angle \mathit{A}\mathit{C}\mathit{B}$ . . . (ii)
In $△\mathit{O}\mathit{B}\mathit{C}$
$\angle 1+\angle 2+\angle \mathit{B}\mathit{O}\mathit{C}={180}^{\circ }$ . . . (iii)
Put Values of Eqs. (i) and (ii) in Eq. (iii) ${90}^{\circ }-\frac{1}{2}\angle \mathit{A}\mathit{B}\mathit{C}+{90}^{\circ }-\frac{1}{2}\angle \mathit{A}\mathit{C}\mathit{B}$
$+\angle \mathit{B}\mathit{O}\mathit{C}={180}^{\circ }$ . . . (iv)
In $△\mathit{A}\mathit{B}\mathit{C},\angle \mathit{A}+\angle \mathit{B}+\angle \mathit{C}={180}^{\circ }$

From Eqs. (iv) and (v),
${180}^{\circ }-\frac{1}{2}\left(\angle \mathit{A}\mathit{B}\mathit{C}+\angle \mathit{A}\mathit{C}\mathit{B}\right)+{63}^{\circ }={180}^{\circ }$
${180}^{\circ }-\frac{1}{2}\left(\angle \mathit{A}\mathit{B}\mathit{C}+\angle \mathit{A}\mathit{C}\mathit{B}\right)+{63}^{\circ }={180}^{\circ }$
$\Rightarrow {180}^{\circ }-\frac{1}{2}\left({180}^{\circ }-\angle \mathit{A}\right)+{63}^{\circ }={180}^{\circ }$
$\Rightarrow \frac{1}{2}\left({180}^{\circ }-\angle \mathit{A}\right)={63}^{\circ }$
$\Rightarrow {90}^{\circ }-\frac{\angle \mathit{A}}{2}={63}^{\circ }$
$\Rightarrow \frac{\angle \mathit{A}}{2}={90}^{\circ }-{63}^{\circ }={27}^{\circ }$
$\therefore \angle \mathit{A}={27}^{\circ }\times 2={54}^{\circ }$