To solve for \(\alpha\) and \(F\), we resolve the forces in the horizontal and vertical directions.

Resolving horizontally:

\(6 \cos \alpha + 5 \cos (90^\circ - \alpha) = F\)

Resolving vertically:

\(6 \sin \alpha - 5 \sin (90^\circ - \alpha) = F\)

Using the identity \(\cos(90^\circ - \alpha) = \sin \alpha\) and \(\sin(90^\circ - \alpha) = \cos \alpha\), we have:

\(6 \cos \alpha + 5 \sin \alpha = 6 \sin \alpha - 5 \cos \alpha\)

Rearranging gives:

\(11 \cos \alpha = \sin \alpha\)

\(\tan \alpha = \frac{1}{11}\)

Solving for \(\alpha\), we find:

\(\alpha = 84.8^\circ\)

Substituting \(\alpha = 84.8^\circ\) back into the horizontal resolution equation:

\(F = 6 \cos 84.8^\circ + 5 \sin 84.8^\circ\)

\(F = 5.52 \text{ N}\)