Question 5Triangle $ABC$ is right-angled at $A$, and $AH$ is its altitude to $BC$. Given $AH=1$ cm and $\angle ABC=60^\circ$, find $BC$ in centimetres. $\frac{2}{\sqrt3}$$2\sqrt2$$2\sqrt3$$\frac4{\sqrt3}$$\frac{2\sqrt2}{\sqrt3}$