Triangle Off The Square

First, we assume the length of the side of the square to be 1. (It will be the same to use any other values, or just $x$ , though.) From the figure above, it will be easy to determine that

$M = \tan^{-1} (A) + \tan^{-1} (B)$

where $A = \frac{\frac {1}{4}}{1 - \frac{\sqrt{3}}{4}}$

$A = \frac{1}{4 - \sqrt{3}}$

$A = \frac{4 + \sqrt{3}}{13}$

and $B = \frac{\frac {3}{4}}{1 - \frac{\sqrt{3}}{4}}$

$B = \frac{3}{4 - \sqrt{3}}$

$B = \frac{3(4 + \sqrt{3})}{13}$

In determining the values of A and B, the fact that the red point creates a 30-60-90 triangle with the square is used. By virtue of similar triangles, the lengths drawn on the figure are determined.

$M = \tan^{-1} (A) + \tan^{-1} (B) \approx \boxed{76.71^\circ}$ .