(i) Resolve the forces horizontally and vertically:

Horizontal component: \(25 \cos 30^{\circ} - 15 \cos 40^{\circ} = 10.1599 \ldots\)

Vertical component: \(25 \sin 30^{\circ} + 15 \sin 40^{\circ} - 30 = -7.8581 \ldots\)

Magnitude of the resultant force:

\(\sqrt{(10.1599)^2 + (-7.8581)^2} = 12.8 \text{ N}\)

Direction: Angle 37.7° below the horizontal in the direction BA.

(ii) For zero component in the direction BC:

\(F \cos 40^{\circ} = 25 \cos 30^{\circ}\)

\(F = 28.3\)

New resultant force:

\(28.26 \sin 40^{\circ} + 25 \sin 30^{\circ} - 30 = 0.667 \text{ N upwards}\)