(i) To find the resultant of the forces in Fig. 1, we calculate the components:

The x-component is given by:

\(x = 4 + 8 \cos 30^\circ + 12 \cos 60^\circ = 10 + 4\sqrt{3} \approx 16.928\)

The y-component is given by:

\(y = 8 \sin 30^\circ + 12 \sin 60^\circ + 16 = 20 + 6\sqrt{3} \approx 30.392\)

The magnitude of the resultant force is:

\(R = \sqrt{x^2 + y^2} = \sqrt{16.928^2 + 30.392^2} \approx 34.8 \text{ N}\)

The direction \(\theta\) with the 4 N force is:

\(\theta = \arctan \left( \frac{y}{x} \right) = \arctan \left( \frac{30.392}{16.928} \right) \approx 60.9^\circ\)

(ii) For Fig. 2, the forces exchange directions. The magnitude remains the same:

\(R = 34.8 \text{ N}\)

The direction \(\theta\) with the 16 N force is:

\(\theta = 90^\circ - 60.9^\circ = 29.1^\circ\)