First, resolve the forces in the x-direction and y-direction:

\(X = 160 + 250 \cos \alpha = 160 + 250 \times 0.96 = 400 \text{ N}\)

\(Y = 370 - 250 \sin \alpha = 370 - 250 \times 0.28 = 300 \text{ N}\)

Calculate the magnitude of the resultant force \(R\) using Pythagoras' theorem:

\(R = \sqrt{X^2 + Y^2} = \sqrt{400^2 + 300^2} = \sqrt{160000 + 90000} = \sqrt{250000} = 500 \text{ N}\)

Calculate the angle \(\theta\) that the resultant makes with the x-direction using the tangent function:

\(\tan \theta = \frac{Y}{X} = \frac{300}{400} = 0.75\)

\(\theta = \arctan(0.75) \approx 36.9^\circ\)

To find the value of \(Q\), we resolve the forces and use trigonometric identities. The angles between the forces and the resultant are given as 40° and 80°.

Using the cosine rule for the triangle formed by the forces and the resultant:

\(Q - P \cos 60^{\circ} = 12 \cos 80^{\circ}\)

\(P \sin 60^{\circ} = 12 \sin 80^{\circ}\)

From the first equation, we can express \(Q\) in terms of \(P\):

\(Q \cos 80^{\circ} + P \cos 40^{\circ} = 12\)

\(P \sin 40^{\circ} = Q \sin 80^{\circ}\)

Eliminating \(P\) from these equations gives:

\(Q \cos 80^{\circ} + \frac{12 \sin 80^{\circ} \cos 60^{\circ}}{\sin 60^{\circ}} = 12 \cos 80^{\circ}\)

Solving for \(Q\):

\(Q = 8.91\)

(a) To find the values of \(F\) and \(G\) when the forces are in equilibrium, resolve the forces vertically and horizontally.

Vertically: \(G \sin 60^\circ + 2F \sin 40^\circ - 10 = 0\)

Horizontally: \(F + G \cos 60^\circ - 2F \cos 40^\circ = 0\)

Solving these equations simultaneously gives \(F = 4.53\) and \(G = 4.82\).

(b) Given \(F = 3\), resolve the forces parallel to the 10N force and equate this expression to zero:

\(G \sin 60^\circ + 2 \times 3 \sin 40^\circ - 10 = 0\)

Solving for \(G\) gives \(G = 7.09\) to 3 significant figures.

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

For the 7 N force along the x-axis:

\(F_{x1} = 7 \text{ N}\)

For the 10 N force at 50° to the x-axis:

\(F_{x2} = 10 \cos 50^{\circ}\)

\(F_{y2} = 10 \sin 50^{\circ}\)

For the 15 N force at 80° to the x-axis:

\(F_{x3} = -15 \cos 80^{\circ}\)

\(F_{y3} = 15 \sin 80^{\circ}\)

The negative sign for \(F_{x3}\) is because the force is acting in the negative x-direction.

Summing the x-components:

\(X = F_{x1} + F_{x2} + F_{x3} = 7 + 10 \cos 50^{\circ} - 15 \cos 80^{\circ}\)

Summing the y-components:

\(Y = F_{y2} + F_{y3} = 10 \sin 50^{\circ} + 15 \sin 80^{\circ}\)

Calculating these gives:

\(X = 10.8 \text{ N}\)

\(Y = 22.4 \text{ N}\)

To find the direction of the resultant force, we use:

\(\theta = \arctan \left( \frac{Y}{X} \right) = \arctan \left( \frac{22.4}{10.8} \right)\)

Calculating this gives:

\(\theta = 64.2^{\circ}\)

The direction is 64.2° anticlockwise from the x-axis.

(i) To find the components of the 8 N force:

1. Parallel to the 10 N force: The component is \(8 \cos \theta\). Therefore, the resultant component parallel to the 10 N force is \(10 - 8 \cos \theta\).
2. Perpendicular to the 10 N force: The component is \(8 \sin \theta\).

(ii) The magnitude of the resultant force is given as 8 N. Using the Pythagorean theorem for the components:

\((10 - 8 \cos \theta)^2 + (8 \sin \theta)^2 = 8^2\)

Expanding and simplifying:

\(100 - 160 \cos \theta + 64 \cos^2 \theta + 64 \sin^2 \theta = 64\)

Using \(\sin^2 \theta + \cos^2 \theta = 1\):

\(100 - 160 \cos \theta + 64 = 64\)

\(164 - 160 \cos \theta = 64\)

\(160 \cos \theta = 100\)

\(\cos \theta = \frac{5}{8}\)