I found: ##(2pi/3)##

Consider the following diagram:

We can see that the relationships between rectangular and are:

##r=sqrt(x^2+y^2)##

##theta=arctan(y/x)##

and:

##x=rcos(theta)##

##y=rsin(theta)##

In your case:

##x=1##

##y=sqrt(3)##

use them into:

##r=sqrt(x^2+y^2)##

##theta=arctan(y/x)##

to get:

##r=sqrt(1+3)=sqrt(4)=2##

##theta=arctan(sqrt(3))=60^@=pi/3rad##