To solve for \(\alpha\) and \(F\), we resolve the forces in the horizontal and vertical directions.

Resolving horizontally:

\(6 \cos \alpha + 5 \cos (90^\circ - \alpha) = F\)

Resolving vertically:

\(6 \sin \alpha - 5 \sin (90^\circ - \alpha) = F\)

Using the identity \(\cos(90^\circ - \alpha) = \sin \alpha\) and \(\sin(90^\circ - \alpha) = \cos \alpha\), we have:

\(6 \cos \alpha + 5 \sin \alpha = 6 \sin \alpha - 5 \cos \alpha\)

Rearranging gives:

\(11 \cos \alpha = \sin \alpha\)

\(\tan \alpha = \frac{1}{11}\)

Solving for \(\alpha\), we find:

\(\alpha = 84.8^\circ\)

Substituting \(\alpha = 84.8^\circ\) back into the horizontal resolution equation:

\(F = 6 \cos 84.8^\circ + 5 \sin 84.8^\circ\)

\(F = 5.52 \text{ N}\)

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

Resolve the forces into horizontal and vertical components:

• Horizontal: \(F \cos \theta = 10\)
• Vertical: \(F \sin \theta = 13\)

Using the tangent function, we have:

\(\tan \theta = \frac{13}{10}\)

Thus, \(\theta = \arctan \left( \frac{13}{10} \right) \approx 52.4^\circ\).

Using Pythagoras' theorem for the resultant force:

\(F^2 = 10^2 + 13^2\)

\(F = \sqrt{10^2 + 13^2} = \sqrt{269} \approx 16.4 \text{ N}\)

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

Resolving forces in the vertical ('j') direction:

\(F \sin 50^\circ = F \sin 20^\circ + 12\)

Solving for \(F\):

\(F (\sin 50^\circ - \sin 20^\circ) = 12\)

\(F = \frac{12}{\sin 50^\circ - \sin 20^\circ} \approx 28.3\)

Resolving forces in the horizontal ('i') direction:

\(G = F \cos 50^\circ + F \cos 20^\circ\)

Substitute \(F = 28.3\):

\(G = 28.3 \cos 50^\circ + 28.3 \cos 20^\circ\)

\(G \approx 44.8\)

To solve for \(F\) and \(\alpha\), we use the equilibrium condition, which states that the sum of the forces in any direction must be zero.

Resolve horizontally:

\(F \cos \alpha - 20 \cos 40 - 100 \sin 20 = 0\)

\(F \cos \alpha = 49.5229 \ldots\)

Resolve vertically:

\(F \sin \alpha + 20 \sin 40 - 100 \cos 20 = 0\)

\(F \sin \alpha = 81.1135 \ldots\)

Now, solve for \(F\):

\(F = \sqrt{(49.5229 \ldots)^2 + (81.1135 \ldots)^2}\)

\(F = 95.0\)

To find \(\alpha\):

\(\tan \alpha = \frac{81.1135 \ldots}{49.5229 \ldots}\)

\(\alpha = 58.6\)

Since the system is in equilibrium, the sum of the forces in any direction is zero. We resolve the forces horizontally and vertically.

For horizontal components:

\(P \cos \theta = (36 - 24) \cos 36.9^{\circ}\)

\(P \cos \theta = 9.6\)

For vertical components:

\(P \sin \theta + 20 = (24 + 36) \sin 36.9^{\circ}\)

\(P \sin \theta + 20 = 36\)

\(P \sin \theta = 16\)

Using the equations:

\(P \cos \theta = 9.6\)

\(P \sin \theta = 16\)

We find \(P\) using:

\(P = \sqrt{16^2 + 9.6^2}\)

\(P = 18.7\)

To find \(\theta\):

\(\theta = \arctan \left( \frac{16}{9.6} \right)\)

\(\theta = 59.0^{\circ}\)