Bài giải:

$\sin \alpha =\frac{1}{2}\Rightarrow {{\sin }^{2}}\alpha =\frac{1}{4}$

Ta có: ${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1\Rightarrow \frac{1}{4}+{{\cos }^{2}}\alpha =1\Rightarrow {{\cos }^{2}}\alpha =\frac{3}{4}$

Vì ${{0}^{o}}<\alpha <{{90}^{o}}$nên $\cos \alpha >0\Rightarrow \cos \alpha =\frac{\sqrt{3}}{2}$

$\tan \alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{1}{2}:\frac{\sqrt{3}}{2}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$