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 \(F\):

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

\(F = 16.5\)