Since the forces are in equilibrium, we can use the cosine rule and Lami's theorem to find the angles between the forces.

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

\(10 \cos \alpha = 8 \quad \text{or} \quad 10 \cos \beta = 6\)

Solving for \(\alpha\) and \(\beta\):

\(\alpha = 36.9^{\circ} \quad \text{or} \quad \beta = 53.1^{\circ}\)

Using Lami's theorem:

\(\frac{10}{\sin 90^{\circ}} = \frac{6}{\sin \gamma} = \frac{8}{\sin \delta}\)

Solving for \(\gamma\) and \(\delta\):

\(\gamma = 143.1^{\circ}\) (angle between 8 N and 10 N)

\(\delta = 126.9^{\circ}\) (angle between 6 N and 10 N)

To find \(\theta\) and \(P\), we resolve the forces horizontally and vertically, and use Lami's theorem.

Step 1: Resolve Vertically

Using vertical equilibrium: \(2P \sin \theta = P \sin 60\)

\(\Rightarrow 2 \sin \theta = \sin 60\)

\(\Rightarrow \theta = 25.7^\circ\)

Step 2: Resolve Horizontally

Using horizontal equilibrium: \(2P \cos \theta + P \cos 60 = 10\)

Substitute \(\theta = 25.7^\circ\):

\(2P \cos 25.7 + P \cdot 0.5 = 10\)

\(\Rightarrow P = 4.34 \text{ N}\)

Alternative Method: Lami's Theorem

Using Lami's theorem:

\(\frac{2P}{\sin 120} = \frac{P}{\sin(180 - \theta)} = \frac{10}{\sin(60 + \theta)}\)

Solve for \(\theta\) and \(P\) using the above relationships, confirming \(\theta = 25.7^\circ\) and \(P = 4.34 \text{ N}\).

To find the values of P and θ, we resolve the forces vertically and horizontally, and use Lami's Theorem.

Vertical Resolution:
The vertical components of the forces must sum to zero for equilibrium:

\(3P \sin 55^{\circ} + P \sin θ + P \sin θ = 20\)

\(3P \sin 55^{\circ} = 20\)

Solving for P:

\(P = \frac{20}{3 \sin 55^{\circ}} \approx 8.14 \text{ N}\)

Horizontal Resolution:
The horizontal components of the forces must also sum to zero:

\(3P \cos 55^{\circ} = 2P \cos θ\)

\(\cos θ = \frac{3 \cos 55^{\circ}}{2}\)

Solving for θ:

\(\cos θ = 1.5 \cos 55^{\circ}\)

\(θ \approx 30.6^{\circ}\)

Alternatively, using Lami's Theorem:

\(\frac{3P}{\sin 90^{\circ}} = \frac{20}{\sin 125^{\circ}}\)

Solving gives:

\(P = 8.14 \text{ N}\)

And for the horizontal forces:

\(\frac{3P}{\sin 90^{\circ}} = \frac{2P \cos θ}{\sin 145^{\circ}}\)

\(\cos θ = 1.5 \sin 145^{\circ}\)

\(θ \approx 30.6^{\circ}\)

To solve for \(F\) and \(\alpha\), we resolve the forces horizontally and vertically, using the fact that the forces are in equilibrium.

First, resolve horizontally:

\(2F + F \cos 60^\circ = 15 \cos \alpha\)

Next, resolve vertically:

\(F \sin 60^\circ = 15 \sin \alpha\)

Using these two equations, we can solve for \(F\) and \(\alpha\). From the mark scheme, we have:

\(F = 5.67\) and \(\alpha = 19.1\)

Alternatively, \(F = \frac{15 \sqrt{7}}{7}\).

To solve for \(P\) and \(\theta\), we resolve the forces horizontally and vertically.

For horizontal equilibrium:

\(12\cos75^\circ + P\cos\theta^\circ = 18\cos65^\circ\)

For vertical equilibrium:

\(18\sin65^\circ + 12\sin75^\circ = 15 + P\sin\theta^\circ\)

We can solve these simultaneous equations by eliminating either \(\theta\) or \(P\).

Using the given mark scheme:

\(P^2 = (18\sin65^\circ + 12\sin75^\circ - 15)^2 + (18\cos65^\circ - 12\cos75^\circ)^2\)

or

\(\theta = \arctan\left(\frac{18\sin65^\circ + 12\sin75^\circ - 15}{18\cos65^\circ - 12\cos75^\circ}\right)\)

Solving these gives:

\(P = 13.7\) or \(\theta = 70.8\)