(i) Since the particle is in equilibrium, the sum of the forces in both the horizontal and vertical directions must be zero.

Resolve the forces horizontally and vertically:

Horizontal: \(7 = F \cos \theta\)

Vertical: \(4 = F \sin \theta\)

Using \(F^2 = 7^2 + 4^2\), we find:

\(F = \sqrt{49 + 16} = \sqrt{65} \approx 8.06\)

Using \(\tan \theta = \frac{4}{7}\), we find:

\(\theta = \arctan\left(\frac{4}{7}\right) \approx 29.7^\circ\)

(ii) When the 7 N force is removed, the resultant force is the vector sum of the remaining forces:

The magnitude is 7 N, and the direction is opposite to the original 7 N force.