SSC CGL 20 Aug 2021 Shift 2 Paper

The angle of triangle are in A.P.
Given, smallest angle is ${50}^{\circ }$ less than largest angle.
We know that,
$\mathit{I}=\mathit{a}+\left(\mathit{n}-1\right)\mathit{d}$
Here, let $\mathit{a}=$ first smallest angle is $\mathit{x}$.
$\mathit{I}=$ largest angle $=\mathit{x}+50$ and $\mathit{n}=3$ (angle) $\Rightarrow 50+\mathit{x}=\mathit{x}+\left(3-1\right)\mathit{d}$
$\Rightarrow 50+\mathit{x}=\mathit{x}+2\mathit{d}$
$\Rightarrow 2\mathit{d}=50$
$\Rightarrow \mathit{d}=25$
The distance between all 3 angle is $25.$
Sum of all 3 triangle is ${180}^{\circ }$.
Then, $\mathit{a}+2$ nd angle $+1={180}^{\circ }$
$\Rightarrow \mathit{x}+\mathit{x}+25+\mathit{x}+50={180}^{\circ }$
$\Rightarrow 3\mathit{x}+75={180}^{\circ }$
$\Rightarrow 3\mathit{x}={105}^{\circ }$
$\Rightarrow \mathit{x}={35}^{\circ }$
$\therefore$ Largest angle $\mathit{I}=\mathit{x}+{50}^{\circ }$
$={35}^{\circ }+{50}^{\circ }={85}^{\circ }$