In triangle $ABC$, $A = 30^\circ$, $B = 45^\circ$ and $a = 10$ cm. Find the length of side $b$.

By the sine rule $\dfrac{b}{\sin B} = \dfrac{a}{\sin A}$, so $b = \dfrac{10\sin 45^\circ}{\sin 30^\circ} = \dfrac{10 \times \frac{\sqrt{2}}{2}}{\frac{1}{2}} = 10\sqrt{2}$ cm. Inverting the ratio gives $\dfrac{10}{\sqrt{2}} = 5\sqrt{2}$.