Let the given trinomials be $P_1(x) = (x - x_1)^2 + y_1$ and $P_2(x) = (x - x_2)^2 + y_2$, where $A_1(x_1; y_1)$ and $A_2(x_2; y_2)$ are the coordinates of the vertices of the parabolas. Then $m(P_1(x)) = y_1$, and $P_1(P_2(x)) = ((x - x_2)^2 + y_2 - x_1)^2 + y_1$. If $y_2 \le x_1$, then the minimal value of the expression $((x - x_2)^2 + y_2 - x_1)^2$ is zero, whence $m(P_1(P_2(x))) - m(P_1(x)) = y_1 - y_1 = 0$. The latter contradicts the fact that $m(P_1(P_2(x))) - m(P_1(x))$ is a positive number. Thus, $y_2 > x_1$, from which it follows that $m(P_1(P_2(x))) = (y_2 - x_1)^2 + y_1$ and $m(P_1(P_2(x))) - m(P_1(x)) = (y_2 - x_1)^2$.

Similarly, $y_1 > x_2$ and $m(P_2(P_1(x))) - m(P_2(x)) = (y_1 - x_2)^2$. Now, the condition of equality of the differences can be rewritten as $(y_1 - x_2)^2 = (y_2 - x_1)^2$. Hence, since $y_2 > x_1$ and $y_1 > x_2$, we obtain $y_1 - x_2 = y_2 - x_1$, that is, $y_2 - y_1 = -(x_2 - x_1)$. Therefore, the desired angle is $45^\circ$.