Problem with finding "x" in triangle

Question

I have got a problem with finding the x. I think the question isn't true or there should more informations on it.

Answer 1

There is certainly enough information. Note that in this answer I will use $\overline{AB}$ to denote the line from $A$ to $B$. Note that you have a 30-60-90 triangle (ABE) which allows you to find ($\overline{BE}$) (you can derive this using trigonometry if you know it, or take a look here. Then, you can also find ($\overline{AE}$) to be

$d=\overline{AE} = \frac{2a}{\sqrt{3}}$

and subsequently ($\overline{ED}$),

$\overline{ED} = \overline{AD}-d$

which gives you one side in a 30-60-90 triangle, allowing you to find ($\overline{EF}$). Now you have two sides and an angle in a triangle, you can easily find $X$ using trigonometry from there!

Answer 2

With coordinates, if $B=(0,0)$, $A=(1,0)$, $D=(1,1)$, $C=(0,1)$, we find that $$E=(1,\frac 1{\sqrt 3})$$ and then $$F=(1-\sqrt3\cdot(1-\frac1{\sqrt 3}),1)=(2-\sqrt 3,1)$$ so that $$ \tan\angle FBC = \sqrt 3-1$$ and finally $$ x = 60^\circ -\arctan(\sqrt 3-1)$$