From the following displacement-time graph

In first instant you will apply $\upsilon = \tan \theta $ and say, $\upsilon = \tan {30^o} = \frac{1}{{\sqrt 3 }}\, m/s.$

But it is wrong because formula $\upsilon = \tan \theta $ is valid when angle is measured with time axis.

Here angle is taken from displacement axis. So angle from time axis $ = {90^o} - {30^o} = {60^o}$

Now $\upsilon = \tan \,{60^o} = \sqrt 3 $