Resolve the forces into their components. For the force of 25 N, the angle with the negative x-axis is tan^-1 0.75, which gives \cos \theta = 0.8 and \sin \theta = 0.6.

The components of the force F are:

\(F_x = F \cos \theta = 25 \times 0.8 = 20\)

\(F_y = F \sin \theta = 63 - 25 \times 0.6 = 48\)

Using equilibrium conditions, \(F = \sqrt{F_x^2 + F_y^2}\) or \(\tan \theta = \frac{F_y}{F_x}\).

Thus, \(F = 52 \, \text{N}\) and \(\tan \theta = 2.4\).