(a) To show \(\left( \frac{14.4}{30 - P} \right)^2 + \left( \frac{28.8}{P + 30} \right)^2 = 1\), resolve forces horizontally and vertically.

Horizontal: \((33 + 15) \times \frac{3}{5} = P \cos \theta + 30 \cos \theta\)

Vertical: \(15 \times \frac{4}{5} + 30 \sin \theta = 33 \times \frac{4}{5} + P \sin \theta\)

Using \(\cos^2 \theta + \sin^2 \theta = 1\), we find:

\(\cos \theta = \frac{28.8}{P + 30}\) and \(\sin \theta = \frac{14.4}{30 - P}\)

Substitute these into \(\cos^2 \theta + \sin^2 \theta = 1\) to get:

\(\left( \frac{14.4}{30 - P} \right)^2 + \left( \frac{28.8}{P + 30} \right)^2 = 1\)

(b) Substitute \(P = 6\):

\(\left( \frac{14.4}{24} \right)^2 + \left( \frac{28.8}{36} \right)^2 = \left( \frac{3}{5} \right)^2 + \left( \frac{4}{5} \right)^2 = 0.36 + 0.64 = 1\)

Thus, \(\theta = 36.9^\circ\).