(a) Resolve the forces horizontally and vertically. For equilibrium:

Horizontal: \(T \sin \theta = 32\)

Vertical: \(T \cos \theta = 80\)

Divide the horizontal equation by the vertical equation:

\(\tan \theta = \frac{32}{80}\)

\(\theta = \arctan \left( \frac{32}{80} \right) \approx 21.8^{\circ}\)

Substitute \(\theta\) back to find \(T\):

\(T = \frac{80}{\cos \theta} \approx 86.2 \text{ N}\)

(b) Given \(T = 120 \text{ N}\), resolve the forces:

Horizontal: \(120 \sin \theta = P\)

Vertical: \(120 \cos \theta = 80\)

Divide the vertical equation by \(T\):

\(\cos \theta = \frac{80}{120}\)

\(\theta = \cos^{-1} \left( \frac{80}{120} \right) \approx 48.2^{\circ}\)

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

\(P = 120 \sin \theta \approx 89.4 \text{ N}\)