(a) To show \(\left( \frac{14.4}{30 - P} \right)^2 + \left( \frac{28.8}{P + 30} \right)^2 = 1\), resolve forces horizontally and vertically.

Horizontal: \((33 + 15) \times \frac{3}{5} = P \cos \theta + 30 \cos \theta\)

Vertical: \(15 \times \frac{4}{5} + 30 \sin \theta = 33 \times \frac{4}{5} + P \sin \theta\)

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

\(\cos \theta = \frac{28.8}{P + 30}\) and \(\sin \theta = \frac{14.4}{30 - P}\)

Substitute these into \(\cos^2 \theta + \sin^2 \theta = 1\) to get:

\(\left( \frac{14.4}{30 - P} \right)^2 + \left( \frac{28.8}{P + 30} \right)^2 = 1\)

(b) Substitute \(P = 6\):

\(\left( \frac{14.4}{24} \right)^2 + \left( \frac{28.8}{36} \right)^2 = \left( \frac{3}{5} \right)^2 + \left( \frac{4}{5} \right)^2 = 0.36 + 0.64 = 1\)

Thus, \(\theta = 36.9^\circ\).

To show that the forces are in equilibrium, we need to resolve the forces horizontally and vertically and show that the net force in each direction is zero.

Resolving horizontally:

\(X = 78 \times \frac{5}{13} - 50 \times \frac{3}{5} = 78 \cos 67.4^\circ - 50 \cos 53.1^\circ\)

\(X = 30 - 30 = 0\)

Resolving vertically:

\(Y = 78 \times \frac{12}{13} + 50 \times \frac{4}{5} - 112 = 78 \sin 67.4^\circ + 50 \sin 53.1^\circ - 112\)

\(Y = 72 + 40 - 112 = 0\)

Since both \(X = 0\) and \(Y = 0\), the forces are in equilibrium.

Alternatively, using Lami's theorem:

\(\frac{112}{\sin 59.5^\circ} = \frac{50}{\sin 157.4^\circ} = \frac{78}{\sin 143.1^\circ}\)

Calculating the exact values:

\(\frac{112}{56/65} = \frac{50}{5/13} = \frac{78}{3/5} = 130\)

Thus, the forces satisfy Lami's theorem, confirming equilibrium.

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

Resolve horizontally:

\(F \cos \alpha = 15 \cos 20^\circ - 5\)

\(F \cos \alpha = 9.095\)

Resolve vertically:

\(F \sin \alpha = 15 \sin 20^\circ + 25\)

\(F \sin \alpha = 30.13\)

Using Pythagoras' theorem to find \(F\):

\(F = \sqrt{(15 \cos 20^\circ - 5)^2 + (15 \sin 20^\circ + 25)^2}\)

\(F = 31.5\)

Using trigonometry to find \(\alpha\):

\(\alpha = \arctan \left( \frac{15 \sin 20^\circ + 25}{15 \cos 20^\circ - 5} \right)\)

\(\alpha = 73.2^\circ\)

To find the single additional force that will produce equilibrium, we need to resolve the given forces into their components and find the resultant force. The additional force will be equal in magnitude and opposite in direction to this resultant force.

Resolve the forces into horizontal (x-axis) and vertical (y-axis) components:

Horizontal component, \(X = 18 + 12\sin 60^\circ - 8\sin 30^\circ = 14 + 6\sqrt{3}\).

Vertical component, \(Y = 8\cos 30^\circ + 12\cos 60^\circ = 6 + 4\sqrt{3}\).

The magnitude of the resultant force is given by:

\(\sqrt{(14 + 6\sqrt{3})^2 + (6 + 4\sqrt{3})^2} = 27.6 \text{ N}\)

The direction of the resultant force is given by:

\(\arctan \left( \frac{6 + 4\sqrt{3}}{14 + 6\sqrt{3}} \right) = 27.9^\circ\)

Thus, the additional force required for equilibrium has a magnitude of 27.6 N and is directed 27.9° below the negative x-axis.

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

Resolving horizontally:

\(3 \cos 60^{\circ} = 2 \cos \theta\)

\(\frac{3}{2} = 2 \cos \theta\)

\(\cos \theta = \frac{3}{4}\)

\(\theta = \cos^{-1}\left(\frac{3}{4}\right) \approx 41.4^{\circ}\)

Resolving vertically:

\(P = 3 \sin 60^{\circ} + 2 \sin \theta\)

\(P = 3 \times \frac{\sqrt{3}}{2} + 2 \times \sin 41.4^{\circ}\)

\(P \approx 3.92 \text{ N}\)