SSC CGL 16 Aug 2015 Shift 1 Paper

∠ROQ = 96°
In ΔOQR
∠OQR + ∠ORQ + ∠QOR = 180°
⇒ $\frac{1}{2}$ ∠PQR + $\frac{1}{2}$ ∠PRQ + 96° = 180°
⇒ $\frac{1}{2}$ (∠PQR + ∠PRQ)
= 180° – 96° = 84°
⇒ ∠PQR + ∠PRQ = 2 × 84°
= 168°
In ΔPQR,
∴ ∠QPR = 180° – 168° = 12°