The diagonals of rectangle $PQRS$ intersect at point $X$.  If $PS = 6$ and $RS=8$, then what is $\sin \angle PXS$?
[asy]

pair P,Q,R,SS,X,F;

SS = (0,0);

P = (0,6);

R = (8,0);

Q= R+P;

X = Q/2;

F = foot(SS,P,R);

draw(F--SS--R--Q--P--SS--Q);

draw(P--R);

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

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

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

label("$S$",SS,SW);

label("$X$",X,S);

label("$F$",F,NE);

draw(rightanglemark(S,F,X,12));

[/asy]

To find $\sin \angle PXS$, we build a right triangle with $\angle PXS$ as one of its acute angles.  We do so by drawing altitude $\overline{SF}$ from $S$ to diagonal $\overline{PR}$ as shown.  We then have $\sin \angle PXS = \sin\angle FXS = \frac{FS}{XS}$.

The Pythagorean Theorem gives us $PR = QS = 10$, so $SX = QS/2 = 5$.  We also have $\triangle FPS \sim \triangle SPR$ by AA Similarity (both are right triangles and $\angle SPR = \angle FPS$), so
\[\frac{FS}{PS} = \frac{SR}{PR}.\]This gives us
\[FS = PS \cdot \frac{SR}{PR} = \frac{6\cdot 8}{10} = \frac{24}{5}.\]Finally, we have \[\sin \angle PXS = \frac{FS}{XS} = \frac{24/5}{5} = \boxed{\frac{24}{25}}.\]