SSC CHSL 4 Mar 2018 Shift 1 Paper

Sum of interior angles of a regular polygon = $\left(\mathit{n}-2\right)\times 180°$
Sum of all interior angles of a pentagon = $\left(5-2\right)\times 180°=540°$
⇒ Measure of an interior angle of a polygon = $\frac{540}{5}=108°$
Sum of all exterior angles of a pentagon = 360°
⇒ Measure of an exterior angle of a polygon = $\frac{360°}{5}=72°$
∴ Required difference = $108°-72°=36°$