Since the forces are in equilibrium, the sum of the forces in both the horizontal and vertical directions must be zero.

Resolve the forces into horizontal and vertical components:

• Horizontal: \(F \cos \theta = 10\)
• Vertical: \(F \sin \theta = 13\)

Using the tangent function, we have:

\(\tan \theta = \frac{13}{10}\)

Thus, \(\theta = \arctan \left( \frac{13}{10} \right) \approx 52.4^\circ\).

Using Pythagoras' theorem for the resultant force:

\(F^2 = 10^2 + 13^2\)

\(F = \sqrt{10^2 + 13^2} = \sqrt{269} \approx 16.4 \text{ N}\)