(a) Resolving forces vertically and horizontally:

Vertical: \(F \sin \theta + 40 \sin 60^\circ - 50 = 0\)

Horizontal: \(F \cos \theta + 10 - 40 \cos 60^\circ = 0\)

From horizontal resolution: \(F \cos \theta = 10\)

From vertical resolution: \(F \sin \theta = 50 - 20\sqrt{3}\)

\(\theta = \arctan(5 - 2\sqrt{3}) \approx 56.9^\circ\)

\(F = \sqrt{(50 - 20\sqrt{3})^2 + 10^2} \approx 18.3 \text{ N}\)

(b) Given \(F = 10\sqrt{2}\) and \(\theta = 45^\circ\):

Vertical component: \(Y = 10\sqrt{2} \sin 45^\circ + 40 \sin 60^\circ - 50 = 20\sqrt{3} - 40\)

Horizontal component: \(X = 10\sqrt{2} \cos 45^\circ + 10 - 40 \cos 60^\circ = 0\)

Resultant force is \(40 - 20\sqrt{3} \text{ N}\) in the same direction as the \(50 \text{ N}\) force.

(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:

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

To find the value of \(F\), we need to consider the horizontal components of the forces, as the resultant is perpendicular to \(F\).

The horizontal component of the \(20 \text{ N}\) force is \(20 \cos 60^\circ\).

The horizontal component of the \(30 \text{ N}\) force is \(30 \cos 60^\circ\).

The horizontal component of the \(F \text{ N}\) force is \(-F\) (since it acts to the left).

Since the resultant is perpendicular to \(F\), the sum of the horizontal components must be zero:

\(20 \cos 60^\circ + 30 \cos 60^\circ - F = 0\)

\(F = 20 \cos 60^\circ + 30 \cos 60^\circ\)

\(F = 20 \times \frac{1}{2} + 30 \times \frac{1}{2}\)

\(F = 10 + 15\)

\(F = 25\)

(i) Resolve the forces horizontally and vertically:

Horizontal component: \(X = 75 + 50 \cos 60^\circ = 100\)

Vertical component: \(Y = 50 \sin 60^\circ = 43.3\)

Resultant force: \(\sqrt{100^2 + 43.3^2} = 109 \text{ N}\)

Direction: \(\text{Angle} = \arctan\left(\frac{43.3}{100}\right) = 23.4^\circ\) anticlockwise from the positive x-axis.

(ii) For equilibrium, resolve forces:

Horizontal: \(50 \cos \alpha - F \cos 50^\circ = 0\)

Vertical: \(50 \sin \alpha - 3F - F \sin 50^\circ = 0\)

\(\tan \alpha = \frac{3F + F \sin 50^\circ}{F \cos 50^\circ}\)

Solving gives \(\alpha = 80.3^\circ\) and \(F = 13.1 \text{ N}\).

(i) To find the total force in the direction of the river, resolve each force into its horizontal component:

\(X = 60 \cos 25^{\circ} + 50 \cos 15^{\circ}\)

Calculating gives:

\(X = 54.49 + 48.30 = 102.79 \text{ N}\)

Rounded to 103 N.

(ii) To find the magnitude and direction of the resultant force, resolve each force into its vertical component:

\(Y = 60 \sin 25^{\circ} - 50 \sin 15^{\circ}\)

Calculating gives:

\(Y = 25.35 - 12.94 = 12.41 \text{ N}\)

Now, use Pythagoras' theorem to find the magnitude:

\(R = \sqrt{X^2 + Y^2} = \sqrt{102.79^2 + 12.41^2}\)

\(R = \sqrt{10568.38 + 154.08} = \sqrt{10722.46} = 103.4 \text{ N}\)

To find the direction, use:

\(\alpha = \arctan\left(\frac{Y}{X}\right) = \arctan\left(\frac{12.41}{102.79}\right)\)

\(\alpha = 6.9^{\circ}\)