Point $B$ is due east of point $A$. Point $C$ is due north of point $B$. The distance between points $A$ and $C$ is $10\sqrt 2$, and $\angle BAC = 45^\circ$. Point $D$ is $20$ meters due north of point $C$. The distance $AD$ is between which two integers?
$\textbf{(A)}\ 30\ \text{and}\ 31 \qquad\textbf{(B)}\ 31\ \text{and}\ 32 \qquad\textbf{(C)}\ 32\ \text{and}\ 33 \qquad\textbf{(D)}\ 33\ \text{and}\ 34 \qquad\textbf{(E)}\ 34\ \text{and}\ 35$
If point B is due east of point A and point C is due north of point B, $\angle CBA$ is a right angle. And if $\angle BAC = 45^\circ$, $\triangle CBA$ is a 45-45-90 triangle. Thus, the lengths of sides $CB$, $BA$, and $AC$ are in the ratio $1:1:\sqrt 2$, and $CB$ is $10 \sqrt 2 \div \sqrt 2 = 10$.
$\triangle DBA$ is clearly a right triangle with $C$ on the side $DB$. $DC$ is 20, so $DB = DC + CB = 20 + 10 = 30$. 
By the Pythagorean Theorem, $DA = \sqrt {DB^2 + BA^2} = \sqrt {30^2 + 10 ^2} = \sqrt {1000}$.
$31^2 = 961$, and $32^2 = 1024$. Thus, $\sqrt {1000}$ must be between $31$ and $32$. The answer is $\boxed {B}$.