AMC 12 A

Because $\alpha$ and $\beta$ are the smallest angles in these triangles, $\sin \alpha = \frac{3}{5}$, $\cos \alpha = \frac{4}{5}$, $\sin \beta = \frac{7}{25}$, and $\cos \beta = \frac{24}{25}$. By a Double Angle Formula,
$$\sin(2\alpha) = 2 \sin \alpha \cdot \cos \alpha = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = \frac{24}{25} = \cos \beta = \sin\left(\frac{\pi}{2} - \beta\right).$$
Because both $2\alpha$ and $\beta$ are acute, $2\alpha = \frac{\pi}{2} - \beta$, so $\beta = \frac{\pi}{2} - 2\alpha$.

Using complex numbers in polar form, $4 + 3i = 5(\cos \alpha + i \sin \alpha)$. Squaring gives $7 + 24i = 25(\cos \alpha + i \sin \alpha)^2$. Similarly, $24 + 7i = 25(\cos \beta + i \sin \beta)$. Multiplying these two equations yields
$$(7 + 24i)(24 + 7i) = 25(\cos \alpha + i \sin \alpha)^2 \cdot 25(\cos \beta + i \sin \beta) \\ 625i = 625 (\cos(2\alpha + \beta) + i \sin(2\alpha + \beta)) \\ \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} = \cos(2\alpha + \beta) + i \sin(2\alpha + \beta).$$
Because both $2\alpha$ and $\beta$ are acute, $2\alpha + \beta = \frac{\pi}{2}$ and $\beta = \frac{\pi}{2} - 2\alpha$.