O is the incentre of ΔPQR and ∠QPR = 50°, then O is the incentre of ΔPQR

the measure of ∠QOR is :

$\angle \mathrm{QPR}=50^{\circ}$

$\therefore \angle \mathrm{PQR}+\angle \mathrm{PRQ}=180^{\circ}-50^{\circ}=130^{\circ}$

$\therefore \frac{1}{2} \angle \mathrm{PQR}+\frac{1}{2} \angle \mathrm{PRQ}=65^{\circ}$

The point of intersection of internal bisectors of angles is in-centre.

$\therefore \quad \angle \mathrm{OQR}=\frac{1}{2} \angle \mathrm{PQR} ;$

$\angle \mathrm{ORQ}=\frac{1}{2} \quad \angle \mathrm{PRQ}$

In $\triangle \mathrm{OQR}, \angle \mathrm{OQR}+\angle \mathrm{QOR}+\angle \mathrm{ORQ}=180^{\circ}$

$\Rightarrow \angle \mathrm{QOR}=180^{\circ}-65^{\circ}=115^{\circ}$