In rectangle $ABCD$, $\overline{AB}=20$ and $\overline{BC}=12$. Let $E$ be a point on $\overline{CD}$ such that $BE=15$. What is $\overline{AE}$? $\textbf{(A)}\ \sqrt{265}\qquad$ $\textbf{(B)}\ 17\qquad$ $\textbf{(C)}\ 10\sqrt{3}\qquad$ $\textbf{(D)}\ 18\qquad$ $\textbf{(E)}\ \sqrt{345}$
Solution
In the right triangle $BCE$, $CE=\sqrt{15^2-12^2}=9$.
$\sin \angle CBE=\frac{CE}{BE}=\frac35$, and $\angle ABE=90^\circ-\angle CBE$, so $\cos \angle ABE=\frac35$.
