SSC CHSL 12 Oct 2020 Shift 1 Paper

Since, $\mathrm{\Delta }\mathrm{AOC}$ is an isoceles triangle.
∴ $\angle \mathrm{OAC}=\angle \mathrm{OCA}=\mathit{x}$ (let)
$\ln \mathrm{\Delta }\mathrm{AOC}$
$\mathit{x}+\mathit{x}+118°=180°$ (by angle sum property of triangle)
$2\mathit{x}=180°-118°$
⇒ $2\mathit{x}=62°$
⇒ $\mathit{x}=31°$
Now, we know that tangent is always perpendicular to the radius, at point of contact,
∴ $\angle \mathrm{OCD}=90°$
Now, $\angle \mathrm{ACD}=\angle \mathrm{OCD}-\angle \mathrm{OCA}$
$\angle \mathrm{ACD}=90°-31°$ $\left(\because \angle \mathrm{OCA}=\mathit{x}=31°\right)$
⇒ $\angle \mathrm{ACD}=59°$