The sides of triangle $PQR$ are tangent to a circle with center $C$ as shown. Given that $\angle PQR = 63^\circ$ and $\angle QPR = 59^\circ$, find $\angle QRC$, in degrees.

[asy]

unitsize(1.0 cm);

pair Q, P, R, C;

Q = (2.43,3.46);

P = (0,0);

R = (4.43,0);

C = incenter(Q,P,R);

draw(Q--P--R--cycle);

draw(incircle(Q,P,R));

draw(R--C);

label("$Q$", Q, N);

label("$P$", P, SW);

label("$R$", R, SE);

label("$C$", C, N);

[/asy]
The circle with center $C$ is the incircle of $\triangle PQR$. So, any segment from a vertex of the triangle to $C$ is an angle bisector.

The sum of the measures of the internal angles of a triangle is $180^\circ$, so

\begin{align*}
\angle QRP &= 180^\circ - \angle PQR - \angle QPR \\
&= 180^\circ - 63^\circ - 59^\circ\\
&= 58^\circ.
\end{align*}Since $\overline{RC}$ bisects $\angle QRP$, we have $\angle QRC = \frac{58^\circ}{2} = \boxed{29^\circ}$.