4. Let the acute angle $\alpha$ satisfy
$$
3 \sqrt{41-40 \cos \alpha}+4 \sqrt{34-30 \sin \alpha}=25 \text {. }
$$

Then $\sin \alpha=$ . $\qquad$
4. $\frac{3}{5}$.

As shown in Figure 4, construct two triangles $\triangle ACD$ and $\triangle BCD$ with a common side $CD$, such that
$$
\begin{array}{l} 
\angle ACD = \alpha, \\
\angle BCD \\
= 90^{\circ} - \alpha, \\
AC = 15, CD = 12, BC = 20.
\end{array}
$$

By the Law of Cosines, we have
$$
\begin{array}{l}
AD = 3 \sqrt{41 - 40 \cos \alpha}, \\
BD = 4 \sqrt{34 - 30 \sin \alpha}.
\end{array}
$$

Thus, $AD + BD = 25$.
Noting that $\angle ACB = 90^{\circ}$, then
$$
AB = \sqrt{AC^2 + BC^2} = 25 = AD + BD.
$$

Therefore, point $D$ lies on $AB$.
It is also easy to see that $CD \perp AB$, so $\angle ABC = \alpha$.
Hence, $\sin \alpha = \frac{AC}{AB} = \frac{3}{5}$.