To solve for \(F\) and \(\theta\), resolve the forces into horizontal and vertical components:

Horizontal: \(F \cos \theta = 12 \cos 30^\circ\)

Vertical: \(F \sin \theta = 10 - 12 \sin 30^\circ\)

Calculate \(F\) and \(\theta\):

\(F \cos \theta = 10.932\)

\(F \sin \theta = 4\)

Using \(F^2 = X^2 + Y^2\):

\(F = \sqrt{10.932^2 + 4^2} = 11.1 \text{ N}\)

Using \(\tan \theta = \frac{Y}{X}\):

\(\theta = \arctan \left( \frac{4}{10.932} \right) = 21.1^\circ\)

For part (ii), when the 12 N force is removed, the resultant force is the vector sum of the remaining forces:

Magnitude is 12 N, direction is 30° clockwise from the positive x-axis.