To solve for \(W_1\) and \(W_2\), we use the triangle of forces and the sine rule. The angles in the triangle are 40°, 60°, and 80°.

Using the sine rule:

\(\frac{W_1}{\sin 60^{\circ}} = \frac{5}{\sin 80^{\circ}}\)

Solving for \(W_1\):

\(W_1 = \frac{5 \cdot \sin 60^{\circ}}{\sin 80^{\circ}} = 4.40 \text{ N}\)

Similarly, for \(W_2\):

\(\frac{W_2}{\sin 40^{\circ}} = \frac{5}{\sin 80^{\circ}}\)

Solving for \(W_2\):

\(W_2 = \frac{5 \cdot \sin 40^{\circ}}{\sin 80^{\circ}} = 3.26 \text{ N}\)