Resolve the forces horizontally:

\(50 \cos 20^\circ + 60 - 100 \sin 30^\circ = 56.984 \ldots\)

Resolve the forces vertically:

\(100 \cos 30^\circ - 50 \sin 20^\circ = 69.501 \ldots\)

Calculate the magnitude of the resultant force \(R\):

\(R = \sqrt{(56.984 \ldots)^2 + (69.501 \ldots)^2}\)

\(R = 89.9 \text{ N}\)

Calculate the angle \(\alpha\):

\(\alpha = \arctan \left( \frac{56.984 \ldots}{69.501 \ldots} \right)\)

\(\alpha = 39.3^\circ\)

(i) To find the components of the resultant force in the x-direction, resolve the forces:

The 24 N force has a component in the x-direction: \(24 \cos 25^\circ\).

The 12 N force has a component in the x-direction: \(-12 \cos 65^\circ\) (negative because it acts in the opposite direction).

Thus, the x-component is:

\(24 \cos 25^\circ - 12 \cos 65^\circ = 16.7 \text{ N}\)

(ii) To find the components of the resultant force in the y-direction, resolve the forces:

The 30 N force acts entirely in the y-direction: \(30 \text{ N}\).

The 24 N force has a component in the y-direction: \(-24 \sin 25^\circ\) (negative because it acts downward).

The 12 N force has a component in the y-direction: \(-12 \sin 65^\circ\) (negative because it acts downward).

Thus, the y-component is:

\(30 - 24 \sin 25^\circ - 12 \sin 65^\circ = 8.98 \text{ N}\)

(iii) To find the direction of the resultant, use the tangent function:

\(\tan \theta = \frac{8.98}{16.7}\)

\(\theta = \arctan \left( \frac{8.98}{16.7} \right) = 28.3^\circ\)

The direction is 28.3° anticlockwise from the x-direction.