For block A, the force triangle involves the weight of A, the normal reaction, and the 18 N force acting parallel to the slope. The angle between the weight and the 18 N force is 30°.

Using the sine rule for the triangle of forces, we have:

\(W_A \sin 30^{\circ} = 18\)

\(W_A = \frac{18}{\sin 30^{\circ}} = 36 \text{ N}\)

For block B, the force triangle involves the weight of B, the normal reaction, and the 18 N force acting horizontally. The angle between the weight and the 18 N force is 30°.

Using the sine rule for the triangle of forces, we have:

\(W_B \sin 30^{\circ} = 18 \cos 30^{\circ}\)

\(W_B = \frac{18 \cos 30^{\circ}}{\sin 30^{\circ}} = 31.2 \text{ N}\)