To solve for \(A\) and \(B\), we can resolve the forces horizontally and vertically at the 25 N weight.

1. Resolve vertically: \(A \cos 30^\circ + B \cos 40^\circ = 25\)

2. Resolve horizontally: \(A \sin 30^\circ = B \sin 40^\circ\)

From the horizontal resolution, we have:

\(A \sin 30^\circ = B \sin 40^\circ\)

\(A \cdot 0.5 = B \cdot 0.6428\)

\(A = \frac{B \cdot 0.6428}{0.5}\)

\(A = 1.2856B\)

Substitute \(A = 1.2856B\) into the vertical resolution equation:

\(1.2856B \cdot 0.8660 + B \cdot 0.7660 = 25\)

\(1.112B + 0.766B = 25\)

\(1.878B = 25\)

\(B = \frac{25}{1.878} \approx 13.3\)

Substitute \(B = 13.3\) back to find \(A\):

\(A = 1.2856 \times 13.3 \approx 17.1\)

Thus, \(A = 17.1\) and \(B = 13.3\).