SSC CGL 16 Aug 2015 Shift 2 Paper

∠ QPR = 50°
∴ ∠ PQR + ∠ PRQ
= 180° – 50° = 130°
∴ $\frac{1}{2}$ ∠ PQR + $\frac{1}{2}$ ∠ PRQ = 65°
The point of intersection of internal bisectors of angles is in-centre.
∴ ∠ OQR = $\frac{1}{2}$ ∠ PRQ ;
∠ ORQ = $\frac{1}{2}$ ∠ PRQ
In Δ OQR
∠ OQR + ∠ QOR + ∠ ORQ
= 180°
⇒ ≤ QOR = 180° - 65° = 115°