(a) Resolving forces vertically and horizontally:

Vertical: \(F \sin \theta + 40 \sin 60^\circ - 50 = 0\)

Horizontal: \(F \cos \theta + 10 - 40 \cos 60^\circ = 0\)

From horizontal resolution: \(F \cos \theta = 10\)

From vertical resolution: \(F \sin \theta = 50 - 20\sqrt{3}\)

\(\theta = \arctan(5 - 2\sqrt{3}) \approx 56.9^\circ\)

\(F = \sqrt{(50 - 20\sqrt{3})^2 + 10^2} \approx 18.3 \text{ N}\)

(b) Given \(F = 10\sqrt{2}\) and \(\theta = 45^\circ\):

Vertical component: \(Y = 10\sqrt{2} \sin 45^\circ + 40 \sin 60^\circ - 50 = 20\sqrt{3} - 40\)

Horizontal component: \(X = 10\sqrt{2} \cos 45^\circ + 10 - 40 \cos 60^\circ = 0\)

Resultant force is \(40 - 20\sqrt{3} \text{ N}\) in the same direction as the \(50 \text{ N}\) force.