SSC CGL 30 Aug 2015 Paper

Sum of interior angles of a pentagon
= (2n – 4) × 90°
= (2 × 5 – 4) × 90° = 540°
If PQ = QR = RS = ST
∴ ∠POQ = ∠QOR = ∠ROS
= ∠SOT = $\frac{180°}{4}$ = 45°
∴ OP = OQ = OR = OS = OT = radii
∴ ∠OPQ = $\frac{180°-45°}{2}$ = $\frac{135°}{2}$
= 4 × $\frac{135°}{2}$ = 270°