The main observation is that R is the center of an excircle of $\triangle APQ$. Also note that because $\measuredangle BAR = \measuredangle CAR$ we have BR=CR s.t. R can also be defined as the intersection of the bisector of $\angle A$ and the bisector of BC.
Let R' be the intersection of the angle bisectors of $\angle QPB$ and $\angle PQC$. Since R' is the center of an excircle of $\triangle APQ$, R' is also on the bisector of $\angle PAQ$.
Let S be on PQ s.t. PS = PB and QS = QC (this is possible because PQ = PB + QC). Now $\triangle PR'S$ is congruent to $\triangle PR'B$ (SAS), so R'B = R'S. Analogously R'C = R'S. So R'B = R'C.
Now we have R' on the bisector of $\angle PAQ$ and R' on the bisector of BC, so R'=R, i.e. R is the center of an excircle of $\triangle APQ$. Hence $\measuredangle PRQ = 180^{\circ} - \measuredangle BPQ - \measuredangle CQP = \frac{1}{2}\measuredangle APQ + \frac{1}{2}\measuredangle AQP = 90^{\circ} - \frac{1}{2}\measuredangle A$.
Justify your answer
from the original post. OP should add what he has tried already. $\endgroup$