To solve for \(\alpha\) and \(F\), we use the equilibrium condition for forces. The forces are in equilibrium, so the sum of the forces in any direction is zero.

Resolve the forces horizontally and vertically:

Horizontal: \(F \cos \alpha = 12\)

Vertical: \(F \sin \alpha + 12 \sin \alpha = 15\)

Using the horizontal equation, \(F = \frac{12}{\cos \alpha}\).

Substitute into the vertical equation:

\(\frac{12 \sin \alpha}{\cos \alpha} + 12 \sin \alpha = 15\)

\(12 \tan \alpha + 12 \sin \alpha = 15\)

\(12 \sin \alpha (1 + \cos \alpha) = 15\)

\(\sin \alpha = \frac{12}{15}\)

\(\alpha = \arcsin \left( \frac{12}{15} \right) = 53.1^\circ\)

Substitute \(\alpha\) back to find \(F\):

\(F = \frac{12}{\cos 53.1^\circ} = 9 \text{ N}\)