Here are a couple of different approaches:
Firstly I’m going to make a copy of triangle ABC, rotated 90 degrees anti-clockwise around point A to form triangle AB’C’.
Since ACEC’ is a 35 by 35 square, we can easily find out what DE and B’E are: 21 and 20 respectively. This forms a Pythagorean triple with the hypotenuse, B’D, equal to 29.
However since BD is also (14+15=) 29, AB equals AB’ and AD is shared by both, triangles ABD and AB’D are similar and angle x is equal to the angle we seek.
But since angle BAB’ = 90 degrees (as B’ was formed by rotating B by 90 degrees), the angle we seek must be half of this, namely 45 degrees.
An alternative approach is to scale down the two parts of the puzzle triangle as below. This can be done because it doesn’t change the angle we are seeking to find.
Similar to before we can rotate the left hand triangle 90 degrees anti-clockwise. Now to complete a rectangle we just need a copy of the 5x2 triangle, which means the triangular space in the middle is a right angled isosceles triangle, so x is 45 degrees. Since x + ? = 90 degrees, ? is also 45 degrees.