Triangle inside two squares 
The point C and B are both in the middle of the square's sides. I need to find the angles of the triangle in the middle. 
I heard from my classmates it is something with tangent, but I'm not so proficient in that area. Please help.
 A: 
See, if you observe carefully, then it's simple
If you are familiar with Analytical Algebra, you just need to find the ratio of sides of the triangle, then use cosine rule,
$$ \dfrac{a^2 - b^2 - c^2}{2bc} = \cos A $$
So, Let's assume side to be $a$, then
$$ AC^2 = (2a)^2 + \left(\dfrac a2\right)^2 \, \text{Why?} $$
Similarly,
$$ BC^2 = (a+ a/2)^2 + (a/2)^2 $$
and $$ AB^2 = a^2 + (a/2)^2 $$
Now, you have sides, use cosine rule to calculate angles ;)
That's it.
A: You should learn about the tangent function (the ratio of the opposite side to the adjacent side of a right-angled triangle) and its inverse the arc-tangent function.
You can then fill in your diagram like this (if the length of a side is $s$)

so by subtraction you can find the angles of the triangle: 


*

*at A: $\frac{\pi}{2} - \arctan\left(\frac12\right)-\arctan\left(\frac14\right)$ in radians  or  $90^\circ - \arctan\left(\frac12\right)-\arctan\left(\frac14\right)$  in degrees

*at B: $\pi- \arctan\left(\frac13\right)-\arctan(2)$ in radians  or  $180^\circ - \arctan\left(\frac13\right)-\arctan(2)$ in degrees

*at C: $\pi- \arctan(4)-\arctan(3)$ in radians  or  $180^\circ - \arctan(4)-\arctan(3)$ in degrees


Rather more sophisticated trigonometry would make these angles $\arctan\left(\frac76\right)$, $\pi +\arctan\left(-7\right)$ and $\arctan\left(\frac7{11}\right)$
A: I feel like we're missing a cute trick here, but here are two ways to proceed that aren't so cute.


*

*It doesn't matter how big the squares are, the angles stay the same, so we can assume that the squares are $2$ by $2.$   Then the segment $AB$ is the hypotenuse of a right triangle with legs $1$ and $2.$  So $AB$ has length $\sqrt{5}$.  Similarly $BC = \sqrt{10}$ and $AC = \sqrt{17}.$   Now using Law of Cosines you can compute all the angles.

*The corner at $A$ is a right angle cut into three pieces.  You want the middle piece.  The piece below has tangent equal to $1/4$ and the piece above has tangent equal to $1/2$.  Find such angles and subtract them from the right angle and you have the angle at $A$.  Do the same for $C$ and $B$ follows easily.
