The midpoint of three edges of a cube are connected, as shown below.  Find $\angle XYZ,$ in degrees

[asy]
unitsize(1.2 cm);

pair A, B, C, D, T, X, Y, Z;
pair x, y, z;

x = (2,-0.2);
y = (1.2,0.8);
z = (0,2);

X = (0,0);
Y = x;
T = y;
A = z;
Z = x + y;
B = x + z;
D = y + z;
C = x + y + z;

draw((C + D)/2--(B + C)/2--(B + Y)/2,red);
draw(X--Y--Z--C--D--A--cycle);
draw(B--A);
draw(B--C);
draw(B--Y);
draw(T--X,dashed);
draw(T--D,dashed);
draw(T--Z,dashed);

label("$X$", (C + D)/2, N);
label("$Y$", (B + C)/2, SE);
label("$Z$", (B + Y)/2, W);
[/asy]
We place the diagram in coordinate space, so that $X = (1,2,2),$ $Y = (2,1,2),$ and $Z = (2,0,1).$  Then $XY = YZ = \sqrt{2}$ and $YZ = \sqrt{6},$ and by the Law of Cosines,
\[\cos \angle XYZ = \frac{XY^2 + YZ^2 - XZ^2}{2 \cdot XY \cdot YZ} = \frac{2 + 2 - 6}{2 \cdot \sqrt{2} \cdot \sqrt{2}} = -\frac{1}{2}.\]Therefore, $\angle XYZ = \boxed{120^\circ}.$

Alternative, we can join midpoints of other edges, as shown below, to form a regular hexagon.  This makes it clear that $\angle XYZ = 120^\circ.$

[asy]
unitsize(1.2 cm);

pair A, B, C, D, T, X, Y, Z;
pair x, y, z;

x = (2,-0.2);
y = (1.2,0.8);
z = (0,2);

X = (0,0);
Y = x;
T = y;
A = z;
Z = x + y;
B = x + z;
D = y + z;
C = x + y + z;

draw((C + D)/2--(B + C)/2--(B + Y)/2--(X + Y)/2,red);
draw((X + Y)/2--(X + T)/2--(D + T)/2--(C + D)/2,red + dashed);
draw(X--Y--Z--C--D--A--cycle);
draw(B--A);
draw(B--C);
draw(B--Y);
draw(T--X,dashed);
draw(T--D,dashed);
draw(T--Z,dashed);

label("$X$", (C + D)/2, N);
label("$Y$", (B + C)/2, SE);
label("$Z$", (B + Y)/2, W);
[/asy]