(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 \(P\):

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

To solve for θ and T, we resolve the forces vertically and horizontally.

Vertically:

\(T \sin \theta + 120 \sin 45^{\circ} = 15g\)

Horizontally:

\(T \cos \theta = 120 \cos 45^{\circ}\)

Using the vertical equation:

\(T \sin \theta = 15g - 120 \sin 45^{\circ}\)

Using the horizontal equation:

\(T = \frac{120 \cos 45^{\circ}}{\cos \theta}\)

Substitute for \(T\) in the vertical equation:

\(\frac{120 \cos 45^{\circ}}{\cos \theta} \sin \theta = 15g - 120 \sin 45^{\circ}\)

Using trigonometric identities and solving for \(\theta\):

\(\tan \theta = \frac{15g - 120 \sin 45^{\circ}}{120 \cos 45^{\circ}}\)

Calculate \(\theta\):

\(\theta = 37.5^{\circ}\)

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

\(T = \sqrt{(15g \sin \theta)^2 + (120 \cos 45^{\circ})^2}\)

Calculate \(T\):

\(T = 107 \text{ N}\)

To solve for \(A\) and \(B\), we can resolve the forces horizontally and vertically at the 25 N weight.

1. Resolve vertically: \(A \cos 30^\circ + B \cos 40^\circ = 25\)

2. Resolve horizontally: \(A \sin 30^\circ = B \sin 40^\circ\)

From the horizontal resolution, we have:

\(A \sin 30^\circ = B \sin 40^\circ\)

\(A \cdot 0.5 = B \cdot 0.6428\)

\(A = \frac{B \cdot 0.6428}{0.5}\)

\(A = 1.2856B\)

Substitute \(A = 1.2856B\) into the vertical resolution equation:

\(1.2856B \cdot 0.8660 + B \cdot 0.7660 = 25\)

\(1.112B + 0.766B = 25\)

\(1.878B = 25\)

\(B = \frac{25}{1.878} \approx 13.3\)

Substitute \(B = 13.3\) back to find \(A\):

\(A = 1.2856 \times 13.3 \approx 17.1\)

Thus, \(A = 17.1\) and \(B = 13.3\).

Resolve the forces vertically and horizontally. Let \(T_A\) and \(T_B\) be the tensions in the strings at angles 20° and 40° respectively.

Vertically, the sum of the vertical components of the tensions must equal the weight of the particle:

\(T_A \sin 20^{\circ} + T_B \sin 40^{\circ} = 16\)

Horizontally, the horizontal components must balance:

\(T_A \cos 20^{\circ} = T_B \cos 40^{\circ}\)

Solving these equations simultaneously:

From the horizontal equation, express \(T_A\) in terms of \(T_B\):

\(T_A = T_B \frac{\cos 40^{\circ}}{\cos 20^{\circ}}\)

Substitute into the vertical equation:

\(T_B \frac{\cos 40^{\circ}}{\cos 20^{\circ}} \sin 20^{\circ} + T_B \sin 40^{\circ} = 16\)

Simplify and solve for \(T_B\):

\(T_B (\frac{\cos 40^{\circ} \sin 20^{\circ}}{\cos 20^{\circ}} + \sin 40^{\circ}) = 16\)

\(T_B = \frac{16}{\frac{\cos 40^{\circ} \sin 20^{\circ}}{\cos 20^{\circ}} + \sin 40^{\circ}} \approx 17.4 \text{ N}\)

Substitute \(T_B\) back to find \(T_A\):

\(T_A = 17.4 \frac{\cos 40^{\circ}}{\cos 20^{\circ}} \approx 14.2 \text{ N}\)

To solve for the tensions in the ropes, we resolve the forces horizontally and vertically.

Let the tension in rope PA be \(T_A\) and in rope PB be \(T_B\).

Resolving horizontally:

\(T_A \cos 50^{\circ} - T_B \cos 10^{\circ} = 0\)

Resolving vertically:

\(T_A \sin 50^{\circ} + T_B \sin 10^{\circ} = 20g\)

Solving these equations simultaneously:

From the horizontal resolution, \(T_A \cos 50^{\circ} = T_B \cos 10^{\circ}\).

Substitute \(T_A = \frac{T_B \cos 10^{\circ}}{\cos 50^{\circ}}\) into the vertical resolution:

\(\frac{T_B \cos 10^{\circ}}{\cos 50^{\circ}} \sin 50^{\circ} + T_B \sin 10^{\circ} = 20g\)

\(T_B \left( \frac{\cos 10^{\circ} \sin 50^{\circ}}{\cos 50^{\circ}} + \sin 10^{\circ} \right) = 20g\)

Solving for \(T_B\):

\(T_B = \frac{20g}{\frac{\cos 10^{\circ} \sin 50^{\circ}}{\cos 50^{\circ}} + \sin 10^{\circ}}\)

\(T_B = 200 \text{ N}\)

Substitute \(T_B\) back to find \(T_A\):

\(T_A = \frac{200 \cos 10^{\circ}}{\cos 50^{\circ}}\)

\(T_A = 306 \text{ N}\)

Alternatively, using Lami's Theorem:

\(\frac{T_A}{\sin 80^{\circ}} = \frac{T_B}{\sin 140^{\circ}} = \frac{20g}{\sin 140^{\circ}}\)

Solving gives the same tensions: \(T_A = 306 \text{ N}\) and \(T_B = 200 \text{ N}\).