Question 1138137
Not the best method, but refering to unit circle,  {{{tan(alpha)=sin(alpha)/cos(alpha)=sqrt(3)}}}  will work for angle {{{pi/3}}}.


Better Method:
Draw right triangle.
vertical leg {{{sqrt(3)}}}, horizontal leg 1.
Find hypotenuse using r,
{{{(sqrt(3))^2+1^2=r^2}}};
{{{r=2}}};


Let {{{alpha}}} be opposite of the vertical leg of {{{sqrt(3)}}}.
{{{tan(alpha)=(sqrt(3)/2)/(1/2)}}}  corresponding to {{{alpha=pi/3}}}.