$\tan A=\frac{1}{\sqrt{3}}$

$\therefore \frac{B C}{A C}=\frac{1}{\sqrt{3}}$

$\therefore \frac{ BC }{1}=\frac{ AC }{\sqrt{3}}=k$ (suppose) (where, $k > 0$ )

$\therefore BC =k$ and $AC =\sqrt{3} k$

But, $m \angle C =90$  $\therefore$ Hypotenuse is $\overline{ AB }$.

$\therefore AB ^{2}= BC ^{2}+ AC ^{2}$

$=k^{2}+(\sqrt{3} k)^{2}=4 k^{2}=(2 k)^{2}$

$\therefore AB =2 k$

Now, $\sin A =\frac{ BC }{ AB }=\frac{k}{2 k}=\frac{1}{2}$

OR

$\tan A=\frac{1}{\sqrt{3}}$

$\therefore A=30$

Now, $ \sin A=\sin 30=\frac{1}{2}$