In triangle $ABC$, $M$ and $N$ are the midpoints of $BC$ and $AC$, respectively.  The perpendicular bisectors of $BC$ and $AC$ intersect at a point $O$ inside the triangle.  If $\angle C = 47^\circ$, then find the measure of $\angle MON$, in degrees.
The angles in quadrilateral $MONC$ add up to $360^\circ$.  Two of these angles, namely $\angle OMC$ and $\angle ONC$, are right.  Therefore, $\angle MON + \angle MCN = 180^\circ$, which means $\angle MON = 180^\circ - \angle MCN = 180^\circ - 47^\circ = \boxed{133^\circ}$.

[asy]
unitsize(3 cm);

pair A, B, C, M, N, O;

O = (0,0);
A = dir(140);
B = dir(200);
C = dir(-20);
M = (B + C)/2;
N = (A + C)/2;

draw(A--B--C--cycle);
draw(M--O--N);

label("$A$", A, dir(90));
label("$B$", B, SW);
label("$C$", C, SE);
label("$M$", M, S);
label("$N$", N, NE);
label("$O$", O, NW);
[/asy]