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

Resolve horizontally:

\(F \cos \alpha = 15 \cos 20^\circ - 5\)

\(F \cos \alpha = 9.095\)

Resolve vertically:

\(F \sin \alpha = 15 \sin 20^\circ + 25\)

\(F \sin \alpha = 30.13\)

Using Pythagoras' theorem to find \(F\):

\(F = \sqrt{(15 \cos 20^\circ - 5)^2 + (15 \sin 20^\circ + 25)^2}\)

\(F = 31.5\)

Using trigonometry to find \(\alpha\):

\(\alpha = \arctan \left( \frac{15 \sin 20^\circ + 25}{15 \cos 20^\circ - 5} \right)\)

\(\alpha = 73.2^\circ\)