Let [tex]x,y[/tex] be the legs of the triangle, with [tex]x[tex]\mathrm{area}_{\rm square} = y^2[/tex]
[tex]\mathrm{area}_{\rm triangle} = \dfrac12 xy[/tex]
The square has 3 times the area of the triangle, so
[tex]y^2 = \dfrac32 xy[/tex]
Meanwhile, in the triangle we have
[tex]\tan(\theta) = \dfrac xy[/tex]
Now,
[tex]y^2 = \dfrac32 xy \implies \dfrac23 = \dfrac xy \implies \boxed{\tan(\theta) = \dfrac23}[/tex]
