To solve for the tensions in the strings, we resolve the forces acting on P horizontally and vertically. Let the tension in PA be T_A and in PB be T_B.

Using the geometry of the triangle formed by the strings and the vertical line, we have:

\(T_A \times \frac{40}{50} + T_B \times \frac{40}{104} = 21\)

Solving for T_A and T_B:

\(T_A \times \frac{30}{50} = T_B \times \frac{96}{104}\)

From the first equation, we find:

\(T_A = 20 \text{ N}\)

Substituting T_A into the second equation gives:

\(T_B = 13 \text{ N}\)

Thus, the tension in AP is 20 N and the tension in BP is 13 N.

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

1. Resolve horizontally:

\(T_1 \cos 35^{\circ} = T_2 \cos 40^{\circ}\)

2. Resolve vertically:

\(T_1 \sin 35^{\circ} + T_2 \sin 40^{\circ} = 2.4g\)

where \(g\) is the acceleration due to gravity, approximately \(9.8 \text{ m/s}^2\).

3. Solve the system of equations:

From the horizontal resolution, express \(T_2\) in terms of \(T_1\):

\(T_2 = \frac{T_1 \cos 35^{\circ}}{\cos 40^{\circ}}\)

Substitute \(T_2\) in the vertical resolution equation:

\(T_1 \sin 35^{\circ} + \left(\frac{T_1 \cos 35^{\circ}}{\cos 40^{\circ}}\right) \sin 40^{\circ} = 2.4 \times 9.8\)

Simplify and solve for \(T_1\):

\(T_1 (\sin 35^{\circ} + \frac{\cos 35^{\circ} \sin 40^{\circ}}{\cos 40^{\circ}}) = 23.52\)

\(T_1 = \frac{23.52}{\sin 35^{\circ} + \frac{\cos 35^{\circ} \sin 40^{\circ}}{\cos 40^{\circ}}} \approx 20.4 \text{ N}\)

4. Substitute \(T_1\) back to find \(T_2\):

\(T_2 = \frac{20.4 \cos 35^{\circ}}{\cos 40^{\circ}} \approx 19.0 \text{ N}\)

To solve for the tensions in the strings, we can use the equilibrium conditions for the forces acting on the particle P. The forces involved are the tensions in the strings S₁ and S₂, and the weight of the particle.

Let the tension in S₁ be T₁ and in S₂ be T₂. The weight of the particle is 21 N acting vertically downward.

Using trigonometry, we can resolve the tensions into horizontal and vertical components. The horizontal components must balance each other, and the sum of the vertical components must balance the weight of the particle.

From the diagram, we can determine the angles using the given lengths:

• For S₁: \\(an^{-1}(0.2/0.52) = 21.8^\circ\\)
• For S₂: \\(an^{-1}(0.2/0.25) = 38.7^\circ\\)

Resolving horizontally:

\(T_1 \cos(21.8^\circ) = T_2 \cos(38.7^\circ)\)

Resolving vertically:

\(T_1 \sin(21.8^\circ) + T_2 \sin(38.7^\circ) = 21\)

Solving these equations simultaneously gives:

\(T_1 = 13\) N

\(T_2 = 20\) N

To find the angle AP₁X, we use the cosine rule in triangle XP₁A:

\(\cos \alpha = \frac{5.5}{7.3}\)

Solving for \(\alpha\), we get \(\alpha = 41.1^\circ\).

For the value of W, we use the sine rule in triangle XP₁C:

\(\frac{W}{\sin(180^\circ - 41.1^\circ)} = \frac{7.3}{\sin 90^\circ}\)

Solving for W, we find \(W = 4.8 \text{ N}\).

To solve for \(W_1\) and \(W_2\), we use the triangle of forces and the sine rule. The angles in the triangle are 40°, 60°, and 80°.

Using the sine rule:

\(\frac{W_1}{\sin 60^{\circ}} = \frac{5}{\sin 80^{\circ}}\)

Solving for \(W_1\):

\(W_1 = \frac{5 \cdot \sin 60^{\circ}}{\sin 80^{\circ}} = 4.40 \text{ N}\)

Similarly, for \(W_2\):

\(\frac{W_2}{\sin 40^{\circ}} = \frac{5}{\sin 80^{\circ}}\)

Solving for \(W_2\):

\(W_2 = \frac{5 \cdot \sin 40^{\circ}}{\sin 80^{\circ}} = 3.26 \text{ N}\)