7.27 In the figure below, $ABCD$ is a square, $M, N$ are the midpoints of $BC$ and $CD$ respectively, then $\sin \theta$ equals
(A) $\frac{\sqrt{5}}{5}$.
(B) $\frac{3}{5}$.
(C) $\frac{\sqrt{10}}{5}$.
(D) $\frac{4}{5}$.
(E) None of the above.
[Solution] Let $\angle B A M=\alpha$, then $\angle D A N=\alpha$. Suppose the side length of the square is 1, then
$$
\begin{array}{c}
A M=\frac{\sqrt{5}}{2} . \\
\therefore \quad \sin \theta=\sin \left(90^{\circ}-2 \alpha\right)=\cos 2 \alpha=\cos ^{2} \alpha-\sin ^{2} \alpha \\
=\frac{A B^{2}-B M^{2}}{A M^{2}}=\frac{1-\left(\frac{1}{2}\right)^{2}}{\left(\frac{\sqrt{5}}{2}\right)^{2}}=\frac{3}{5} .
\end{array}
$$

Therefore, the answer is $(B)$.