To find the resultant force, resolve the forces horizontally and vertically.

For the horizontal component:

\(X = 7 - 8 \cos \alpha - 6 \sin \alpha\)

Given \(\sin \alpha = \frac{3}{5}\), use \(\cos \alpha = \frac{4}{5}\) (since \(\sin^2 \alpha + \cos^2 \alpha = 1\)).

\(X = 7 - 8 \times \frac{4}{5} - 6 \times \frac{3}{5} = -3\)

For the vertical component:

\(Y = 8 \sin \alpha - 6 \cos \alpha\)

\(Y = 8 \times \frac{3}{5} - 6 \times \frac{4}{5} = 0\)

The resultant force is \(\sqrt{X^2 + Y^2} = \sqrt{(-3)^2 + 0^2} = 3 \text{ N}\).

Since \(X = -3\), the direction is to the left.

(i) Resolve the forces horizontally and vertically:

Horizontal component: \(R_x = 40 \times \frac{24}{25} - 30 \times \frac{7}{25} = 30\)

Vertical component: \(R_y = 50 - 40 \times \frac{7}{25} - 30 \times \frac{24}{25} = 10\)

Magnitude of the resultant force:

\(R = \sqrt{R_x^2 + R_y^2} = \sqrt{30^2 + 10^2} = 31.6 \text{ N}\)

Direction of the resultant force:

\(\theta = \arctan \left( \frac{R_y}{R_x} \right) = \arctan \left( \frac{10}{30} \right) = 18.4^\circ\)

(ii) To make the resultant act in the positive \(x\)-direction, set \(R_y = 0\):

\(50 - P \times \frac{7}{25} - 30 \times \frac{24}{25} = 0\)

Solving for \(P\):

\(P = 40\)

Resolve the forces in the horizontal direction:

\(120 \cos 75^{\circ} = 150 - 100 - P \cos \theta\)

\(120 \cos 75^{\circ} = 50 - P \cos \theta\)

Resolve the forces in the vertical direction:

\(120 \sin 75^{\circ} = P \sin \theta\)

Using the Pythagorean identity for the resultant force:

\(P^2 = (P \cos \theta)^2 + (P \sin \theta)^2\)

\(P^2 = 14400 - 12000 \cos 75^{\circ} + 2500\)

Alternatively, use the tangent identity:

\(\tan \theta = \frac{120 \sin 75^{\circ}}{50 - 120 \cos 75^{\circ}}\)

Solving these equations gives:

\(P = 117\) N

\(\theta = 80.7^{\circ}\)

(i) To find the resultant of the forces in Fig. 1, we calculate the components:

The x-component is given by:

\(x = 4 + 8 \cos 30^\circ + 12 \cos 60^\circ = 10 + 4\sqrt{3} \approx 16.928\)

The y-component is given by:

\(y = 8 \sin 30^\circ + 12 \sin 60^\circ + 16 = 20 + 6\sqrt{3} \approx 30.392\)

The magnitude of the resultant force is:

\(R = \sqrt{x^2 + y^2} = \sqrt{16.928^2 + 30.392^2} \approx 34.8 \text{ N}\)

The direction \(\theta\) with the 4 N force is:

\(\theta = \arctan \left( \frac{y}{x} \right) = \arctan \left( \frac{30.392}{16.928} \right) \approx 60.9^\circ\)

(ii) For Fig. 2, the forces exchange directions. The magnitude remains the same:

\(R = 34.8 \text{ N}\)

The direction \(\theta\) with the 16 N force is:

\(\theta = 90^\circ - 60.9^\circ = 29.1^\circ\)

To find the resultant of the forces, we resolve each force into its x and y components.

For the x-direction:

\(X = 5 - 7\cos 60^\circ - 3\cos 30^\circ\)

\(X = 5 - 7 \times 0.5 - 3 \times \frac{\sqrt{3}}{2}\)

\(X = 5 - 3.5 - 2.598 \approx -1.098\)

For the y-direction:

\(Y = 7\sin 60^\circ - 3\sin 30^\circ - 4\)

\(Y = 7 \times \frac{\sqrt{3}}{2} - 3 \times 0.5 - 4\)

\(Y = 6.062 - 1.5 - 4 \approx 0.562\)

The magnitude of the resultant force \(R\) is given by:

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

\(R = \sqrt{(-1.098)^2 + (0.562)^2}\)

\(R = \sqrt{1.206 + 0.316} \approx 1.23 \text{ N}\)

The direction \(\theta\) of the resultant force is given by:

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

\(\theta = \arctan\left(\frac{0.562}{-1.098}\right)\)

\(\theta \approx 152.9^\circ \text{ anticlockwise from the positive x-axis}\)