The triangle formed by the lines given by $\sqrt{3}x+y=0,$ $\sqrt{3}x-y=0,$and  $y-2=0$

$\text{ The line }y-2=0\text{ is a horizontal line. }$

$\text{ The slope of the line }\sqrt{3}x+y=0\text{ is }m=-\sqrt{3}\text{ and the slope of the line }\sqrt{3}x-y=0,\text{ is }m=\sqrt{3}$

$\text{ Each acute angle made by these two lines with horizontal line is }{60}^{\circ }$

It means two angles in a triangle is $60°$, therefore, the triangle formed by the given three lines is equilateral triangle.