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

Given \(\tan \alpha = \frac{7}{24}\), we have \(\cos \alpha = \frac{24}{25}\) and \(\sin \alpha = \frac{7}{25}\).

Resolving horizontally:

\(P \cos \theta = 48 \cos \alpha - 14 \sin \alpha\)

\(P \cos \theta = 48 \left(\frac{24}{25}\right) - 14 \left(\frac{7}{25}\right) = 42.16\)

Resolving vertically:

\(P \sin \theta = 50 - 48 \sin \alpha - 14 \cos \alpha\)

\(P \sin \theta = 50 - 48 \left(\frac{7}{25}\right) - 14 \left(\frac{24}{25}\right) = 23.12\)

Using Pythagoras' theorem:

\(P = \sqrt{42.16^2 + 23.12^2} = 48.1\)

To find \(\theta\):

\(\tan \theta = \frac{23.12}{42.16}\)

\(\theta = 28.7^\circ\)

Since the forces are in equilibrium, the sum of the horizontal and vertical components must be zero.

Resolving horizontally:

\(P \cos 25^\circ = 22 + 16 \cos 55^\circ\)

Resolving vertically:

\(Q + 16 \sin 55^\circ = P \sin 25^\circ\)

First, solve the horizontal equation for \(P\):

\(P = \frac{22 + 16 \cos 55^\circ}{\cos 25^\circ}\)

Substitute \(P\) into the vertical equation:

\(Q + 16 \sin 55^\circ = \left(\frac{22 + 16 \cos 55^\circ}{\cos 25^\circ}\right) \sin 25^\circ\)

Solve for \(Q\):

\(Q = \left(\frac{22 + 16 \cos 55^\circ}{\cos 25^\circ}\right) \sin 25^\circ - 16 \sin 55^\circ\)

Calculating these values gives:

\(P \approx 34.4\)

\(Q \approx 1.43\)

(i) To find F, resolve the forces in the horizontal direction. The equilibrium condition gives:

\(15 + F \cos 60^\circ = F \cos 30^\circ\)

Using \(\cos 60^\circ = \frac{1}{2}\) and \(\cos 30^\circ = \frac{\sqrt{3}}{2}\), the equation becomes:

\(15 + \frac{F}{2} = \frac{F\sqrt{3}}{2}\)

Rearranging gives:

\(15 = \frac{F\sqrt{3}}{2} - \frac{F}{2}\)

\(15 = \frac{F(\sqrt{3} - 1)}{2}\)

Solving for \(F\):

\(F = \frac{30}{\sqrt{3} - 1}\)

Rationalizing the denominator:

\(F = \frac{30(\sqrt{3} + 1)}{2} = 15(\sqrt{3} + 1)\)

Calculating gives \(F \approx 41.0\).

(ii) To find G, resolve the forces in the vertical direction. The equilibrium condition gives:

\(G = F(\sin 30^\circ + \sin 60^\circ)\)

Using \(\sin 30^\circ = \frac{1}{2}\) and \(\sin 60^\circ = \frac{\sqrt{3}}{2}\), the equation becomes:

\(G = F\left(\frac{1}{2} + \frac{\sqrt{3}}{2}\right)\)

\(G = F\left(\frac{1 + \sqrt{3}}{2}\right)\)

Substituting \(F = 41.0\):

\(G = 41.0 \times \frac{1 + \sqrt{3}}{2}\)

Calculating gives \(G \approx 56.0\).

Resolve the forces into their components. For the force of 25 N, the angle with the negative x-axis is tan^-1 0.75, which gives \cos \theta = 0.8 and \sin \theta = 0.6.

The components of the force F are:

\(F_x = F \cos \theta = 25 \times 0.8 = 20\)

\(F_y = F \sin \theta = 63 - 25 \times 0.6 = 48\)

Using equilibrium conditions, \(F = \sqrt{F_x^2 + F_y^2}\) or \(\tan \theta = \frac{F_y}{F_x}\).

Thus, \(F = 52 \, \text{N}\) and \(\tan \theta = 2.4\).

To solve for \(\alpha\) and \(F\), we use the equilibrium condition for forces. The forces are in equilibrium, so the sum of the forces in any direction is zero.

Resolve the forces horizontally and vertically:

Horizontal: \(F \cos \alpha = 12\)

Vertical: \(F \sin \alpha + 12 \sin \alpha = 15\)

Using the horizontal equation, \(F = \frac{12}{\cos \alpha}\).

Substitute into the vertical equation:

\(\frac{12 \sin \alpha}{\cos \alpha} + 12 \sin \alpha = 15\)

\(12 \tan \alpha + 12 \sin \alpha = 15\)

\(12 \sin \alpha (1 + \cos \alpha) = 15\)

\(\sin \alpha = \frac{12}{15}\)

\(\alpha = \arcsin \left( \frac{12}{15} \right) = 53.1^\circ\)

Substitute \(\alpha\) back to find \(F\):

\(F = \frac{12}{\cos 53.1^\circ} = 9 \text{ N}\)