Resolving forces horizontally, we have:

\(20 \cos \theta = 4P \cos 30\)

\(20 \cos \theta = 4P \times \frac{\sqrt{3}}{2}\)

\(\cos \theta = \frac{\sqrt{3}}{10} P\)

Resolving forces vertically, we have:

\(4P + 2P \sin 30 = 20 \sin \theta\)

\(4P + P = 20 \sin \theta\)

\(\sin \theta = \frac{P}{4}\)

Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):

\(\left( \frac{\sqrt{3}}{10} P \right)^2 + \left( \frac{P}{4} \right)^2 = 1\)

\(\frac{3}{100} P^2 + \frac{1}{16} P^2 = 1\)

Solving for \(P\), we find \(P = 3.29\).

Substituting \(P\) back to find \(\theta\):

\(\cos \theta = \frac{\sqrt{3}}{10} \times 3.29\)

\(\theta = 55.3\)