Let $BD$ be a median in $\triangle{ABC}$. If $AB=\frac{4\sqrt{6}}{3}$, $\cos{B}=\dfrac{\sqrt{6}}{6}$, and $BD=\sqrt{5}$, find the length of $BC$ and the value of $\sin{A}$.
We have $DE=\frac{1}{2}\cdot AB=\frac{2\sqrt{3}}{3}$. Now consider $\triangle{BDE}$, by the Law of Consines:
$$BD^2 = DE^2 +BE^2-2\cdot BE\cdot ED\cdot \cos{\angle{BED}}$$
Because $AB\parallel ED$, therefore $\angle{BED}=\pi-\angle{B}$. It follows
$$BD^2 = DE^2 +BE^2-2\cdot BE\cdot ED\cdot \cos{\pi-\angle{BED}}$$or
$$(\sqrt{5})^2 = (\frac{\sqrt{6}}{2})^2 +(\frac{1}{2}BC)^2-2\cdot (\frac{1}{2}BC)\cdot (\frac{2\sqrt{6}}{2})\cdot (-\frac{\sqrt{6}}{6})$$
Solving the above equation leads to $\boxed{BC=2}$.
Now consider $\triangle{ABC}$, apply the Law of Consines againe:
$$AC^2 = AB^2 + BC^2 - 2\cdot AB\cdot BC \cdot \cos{B}=\frac{28}{3}\implies AC=\frac{2\sqrt{21}}{3}$$
By the Law of Sines:
$$\frac{BC}{\sin{A}}=\frac{AC}{\sin{B}}\implies \boxed{\sin{A}=\frac{\sqrt{70}}{14}}$$