Finding the sum of two angles I gotta find the value of $x+y$ in the following image

I have no info about if a point is the middle point of a length or even the figure is a square. I can prove that
$$x-y=30°$$
And I tried to use similarity between triangles (and so, parallelism)  but I got no clue at all.
 A: I'll first assume that the outer shape is a rectangle (since all angles are $90$ degrees, I just need confirmation that the lines are straight). I'll be coordinate bashing and referring to the following throughout my solution: https://www.desmos.com/calculator/tufrmfsip0.
Suppose the rectangle has length and width $w,h$ such that the vertices are $(0,0),(w,0),(w,h),(0,h)$. Then the black line (the first edge of the inner right angle triangle) will have gradient $\tan(60)=\sqrt{3}$ and thus it has the equation $y=\sqrt{3}x$. This line meets the square again at $(\frac{h}{\sqrt{3}},h)$. The second edge of the inner right angle triangle (the red line), has the form $-\frac{1}{\sqrt{3}}x+c$ since it is perpendicular to the black line. You can easily obtain that $c=\frac{4h}{3}$ since it passes through $(\frac{h}{\sqrt{3}},h)$. It will meet the square again at $(w,-\frac{w}{\sqrt{3}}+\frac{4h}{3})$.
Now we can obtain that:
$$\tan(x)=\frac{w}{-\frac{w}{\sqrt{3}}+\frac{4h}{3}}$$
To which is follows that since $x-y=30$, then:
$$x+y=2\tan^{-1}(\frac{w}{-\frac{w}{\sqrt{3}}+\frac{4h}{3}})-30$$
Please note that all trigonometry above uses degrees (not radians). I hope this helps and happy to clarify anything!
A: Well I can prove that you can't find the answer without knowing if the figure is a square...Since you have the value for $x-y$ you need $2y$ so that you could add them up and get $x+y$...Even if you did something else and got $x+y$ what your essentially doing is finding $2y$ and adding it to $x-y$...Now think about the figure. Imagine the square or the rectangle being a bit longer in the horizontal direction...Then the contact point that makes the angle $x$, (contact point of the the right angle triangle that has $y$, with the square or the rectangle)  will be further away which in turn will increase the value of $y$ and $x$ equally (thats why $x-y$ is constant).
But since that increases $y$, $2y$ can't have a fixed value...And $x$ doesn't cancel out that increase in value of $2y$ cuz $x$ also increases
Sorry it looks like your question doesn't give you enough details...maybe its wrong...
I hope my proof proved that you can't possibly find the answer to this without knowing if thats a square or a rectangle
