smc

$AB$ is the side of length $10$, and $CD$ is the median of length $9$. The altitude of $C$ to $AB$ is $6$ because the 0.5(altitude)(base)=Area of the triangle. $\theta$ is $\angle CDE$. To find $\sin \theta$, just use opposite over hypotenuse with the right triangle $△DCE$. This is equal to $\frac{6}{9}=\boxed{\frac{2}{3}}$.