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

For the vertical direction:

\(F \sin \theta + 20 \sin 60^\circ - 30 \sin \alpha - 40 \sin \beta = 0\)

Substituting the given values:

\(F \sin \theta + 20 \times 0.866 - 30 \times 0.28 - 40 \times 0.6 = 0\)

\(F \sin \theta = 15.07949 \ldots\)

For the horizontal direction:

\(F \cos \theta + 40 \times 0.8 - 30 \times 0.96 - 20 \cos 60^\circ = 0\)

\(F \cos \theta = 6.8\)

Now, find \(\theta\):

\(\theta = \arctan \left( \frac{15.0794 \ldots}{6.8} \right)\)

\(\theta = 65.7^\circ\)

Finally, find \(F\):

\(F = \sqrt{15.07949^2 + 6.8^2}\)

\(F = 16.5\)

Resolve forces horizontally:

\(P \cos \theta = 12 + 8 \cos 30^\circ - 10 \cos 45^\circ\)

\(P \cos \theta = 11.857\)

Resolve forces vertically:

\(P \sin \theta = 10 \sin 45^\circ - 8 \sin 30^\circ\)

\(P \sin \theta = 3.071\)

Calculate \(P\):

\(P = \sqrt{11.857^2 + 3.071^2}\)

\(P = 12.2\)

Calculate \(\theta\):

\(\theta = \arctan \left( \frac{3.071}{11.857} \right)\)

\(\theta = 14.5\)

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

Resolving horizontally:

\(F \cos \alpha = 40 \sin 30^\circ + 20 \sin 60^\circ - 50 \sin 45^\circ\)

\(F \cos \alpha = 1.965 \ldots\)

Resolving vertically:

\(F \sin \alpha = 50 \cos 45^\circ + 20 \cos 60^\circ - 40 \cos 30^\circ\)

\(F \sin \alpha = 10.714 \ldots\)

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

\(F = \sqrt{(1.965 \ldots)^2 + (10.714 \ldots)^2} = 10.9 \text{ N}\)

To find \(\alpha\):

\(\alpha = \arctan\left(\frac{10.714 \ldots}{1.965 \ldots}\right) = 79.6^\circ\)

Resolving forces horizontally, we have:

\(20 \cos \theta = 4P \cos 30\)

\(20 \cos \theta = 4P \times \frac{\sqrt{3}}{2}\)

\(\cos \theta = \frac{\sqrt{3}}{10} P\)

Resolving forces vertically, we have:

\(4P + 2P \sin 30 = 20 \sin \theta\)

\(4P + P = 20 \sin \theta\)

\(\sin \theta = \frac{P}{4}\)

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

\(\left( \frac{\sqrt{3}}{10} P \right)^2 + \left( \frac{P}{4} \right)^2 = 1\)

\(\frac{3}{100} P^2 + \frac{1}{16} P^2 = 1\)

Solving for \(P\), we find \(P = 3.29\).

Substituting \(P\) back to find \(\theta\):

\(\cos \theta = \frac{\sqrt{3}}{10} \times 3.29\)

\(\theta = 55.3\)

To solve for P and \(\theta\), we resolve the forces horizontally and vertically, as the system is in equilibrium.

Resolving horizontally:

\(P \cos \theta = 32 \cos 20^\circ - 17 \sin 55^\circ\)

\(P \cos \theta = 16.1446\)

Resolving vertically:

\(P \sin \theta = 40 + 17 \cos 55^\circ - 32 \sin 20^\circ\)

\(P \sin \theta = 38.8062\)

Using Pythagoras' theorem to find P:

\(P = \sqrt{(17 \cos 55^\circ - 32 \sin 20^\circ + 40)^2 + (32 \cos 20^\circ - 17 \cos 35^\circ)^2}\)

\(P = 42.0\)

Using trigonometry to find \(\theta\):

\(\theta = \arctan \left( \frac{17 \cos 55^\circ - 32 \sin 20^\circ + 40}{32 \cos 20^\circ - 17 \cos 35^\circ} \right)\)

\(\theta = 67.4\)