$ABCD$ is a square and $M$ and $N$ are the midpoints of $BC$ and $CD$ respectively. Then $\sin \theta=$

$\textbf{(A)}\ \frac{\sqrt{5}}{5} \qquad \textbf{(B)}\ \frac{3}{5} \qquad \textbf{(C)}\ \frac{\sqrt{10}}{5} \qquad \textbf{(D)}\ \frac{4}{5}\qquad \textbf{(E)}\ \text{none of these}$
Use the Sine Area Formula. We can isolate the triangle for which the angle $\theta$ is contained in. WLOG, denote the side length of a triangle as $2$. Our midpoints are then $1$. Subtract the areas of the triangles that don't include the area of our desired triangle: $4-1-1-\frac{1}{2} = \frac{3}{2}.$ The Sine Area Formula tells us $\frac{1}{2}(\sqrt{5})^2\sin\theta=\frac{3}{2}.$ Solving this equation, we get $\sin\theta=\textbf{(B)}\ \frac{3}{5} \qquad$
Solution: Everyoneintexas