Measure the angle!

Extend $CR$ to point $D$ such that $RD=RB$ . Draw $BD$ .

Applying Menelaus' Theorem on $\bigtriangleup BAQ$ with $RPC$ being the transversal, we get,

$\frac{BR}{RA}$ $\cdot$ $\frac{AC}{CQ}$ $\cdot$ $\frac{QP}{PB}$ $=$ $1$

$\Rightarrow$ $\frac{\bcancel{BR}}{\bcancel{RA}}$ $\cdot$ $\frac{AC}{\bcancel{CQ}}$ $\cdot$ $\frac{\bcancel{QP}}{PB}$ $=$ $1$ [ Since $BR=RA$ & $CQ=PQ$ ]
$\Rightarrow$ $BP=AC$

Observe that, in $\bigtriangleup BPD$ & $\bigtriangleup ACR$ , $PD= PR+RD=PR+CP=CR$ , $\angle DPB=\angle RCA$ & $BP=AC$ . Hence, $\bigtriangleup BPD$ $\cong$ $\bigtriangleup ACR$ by $S-S-A$ criterion of congruence.

$\Rightarrow$ $BD=AR=RB$

In $\bigtriangleup BDR$ , $BD=RB=DR$ ; Hence, $\bigtriangleup BDR$ is equilateral.

Therefore, $\angle BRC=180-\angle DRB=180-60=\boxed {120^{\circ}}$