Question:
$AB = 20$ cm, $m \angle A = 30^\circ$, and $m \angle C = 45^\circ$. Express the number of centimeters in the length of $\overline{BC}$ in simplest radical form. [asy]
import olympiad; size(200); import geometry; import graph; defaultpen(linewidth(0.8));
pair A = origin, B = (10*sqrt(3),10), C = (10*sqrt(3) + 10,0);
draw(Label("$20$",align=NW),A--B); draw(B--C); draw(A--C);
label("$A$",A,W); label("$B$",B,N); label("$C$",C,E);
[/asy]

Answer:
Drawing altitude $\overline{BD}$ splits $\triangle ABC$ into 30-60-90 triangle $ABD$ and 45-45-90 triangle $BCD$:

[asy]
import olympiad; size(200); import geometry; import graph; defaultpen(linewidth(0.8));
pair A = origin, B = (10*sqrt(3),10), C = (10*sqrt(3) + 10,0);
draw(Label("$20$",align=NW),A--B); draw(B--C); draw(A--C);
label("$A$",A,W); label("$B$",B,N); label("$C$",C,E);
pair D = (10*sqrt(3),0);
label("$D$",D,S);
draw(B--D);
draw(rightanglemark(B,D,A,40));
[/asy]

From 30-60-90 triangle $ABD$, we have $BD = AB/2 = 10$.  From 45-45-90 triangle $BCD$, we have $BC = BD\sqrt{2} = \boxed{10\sqrt{2}}$.