Sides $AB,~ BC$, and $CD$ of (simple*) quadrilateral $ABCD$ have lengths $4,~ 5$, and $20$, respectively. 
If vertex angles $B$ and $C$ are obtuse and $\sin C = - \cos B =\frac{3}{5}$, then side $AD$ has length

A polygon is called “simple” if it is not self intersecting.
$\textbf{(A) }24\qquad \textbf{(B) }24.5\qquad \textbf{(C) }24.6\qquad \textbf{(D) }24.8\qquad \textbf{(E) }25$
We know that $\sin(C)=-\cos(B)=\frac{3}{5}$. Since $B$ and $C$ are obtuse, we have $\sin(180-C)=\cos(180-B)=\frac{3}{5}$. It is known that $\sin(x)=\cos(90-x)$, so $180-C=90-(180-C)=180-B$. We simplify this as follows:
\[-90+C=180-B\]
\[B+C=270^{\circ}\]
Since $B+C=270^{\circ}$, we know that $A+D=360-(B+C)=90^{\circ}$. Now extend $AB$ and $CD$ as follows:

Let $AB$ and $CD$ intersect at $E$. We know that $\angle AED=90^{\circ}$ because $\angle E = 180 - (A+D)=180-90 = 90^{\circ}$.
Since $\sin BCD = \frac{3}{5}$, we get $\sin ECB=\sin(180-BCD)=\sin BCD = \frac{3}{5}$. Thus, $EB=3$ and $EC=4$ from simple sin application.
$AD$ is the hypotenuse of right $\triangle AED$, with leg lengths $AB+BE=7$ and $EC+CD=24$. Thus, $AD=\boxed{\textbf{(E)}25}$
-WannabeCharmander