In triangle ABC, if the median AD makes an angle $\theta$  with BC and $AB=2$ AD then $\sin \theta =$

$\begin{array}{l}AB=2AD\Rightarrow AD=\frac{AB}{2}\\ \Rightarrow AD=\frac{C}{2}\end{array}$

$W.K.T.$That a median divides a triangle in to two equal areas.

$\begin{array}{l}\therefore \left[2\Delta ADC\right]:\left[\Delta ABC\right]\Rightarrow 2\left(\frac{1}{2}.\frac{C}{2}.\frac{a}{2}\sin \theta \right)\\ =\frac{1}{2}ac\sin B\\ \Rightarrow \sin \theta =2\sin B\end{array}$