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 \(F = 28.3\):

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

\(G \approx 44.8\)