Question

Question on the image​

Answer 1

Let $x,y$ be the legs of the triangle, with $x[tex]\mathrm{area}_{\rm square} = y^2$

$\mathrm{area}_{\rm triangle} = \dfrac12 xy$

The square has 3 times the area of the triangle, so

$y^2 = \dfrac32 xy$

Meanwhile, in the triangle we have

$\tan(\theta) = \dfrac xy$

Now,

$y^2 = \dfrac32 xy \implies \dfrac23 = \dfrac xy \implies \boxed{\tan(\theta) = \dfrac23}$