Resolve the forces horizontally:

\(20 \cos 60^\circ = T \cos 45^\circ\)

\(T = \frac{20 \cos 60^\circ}{\cos 45^\circ} = 10\sqrt{2}\) or 14.1 \text{ N}

Resolve the forces vertically:

\(20 \sin 60^\circ + T \sin 45^\circ = mg\)

\(m = \frac{20 \sin 60^\circ + T \sin 45^\circ}{g}\)

Substitute \(T = 10\sqrt{2}\):

\(m = \frac{20 \times \frac{\sqrt{3}}{2} + 10\sqrt{2} \times \frac{\sqrt{2}}{2}}{g} = 2.73 \text{ kg}\)

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

Vertically, the equilibrium condition gives:

\(T_A \times \frac{4}{5} + T_B \times \frac{3}{5} = 5\)

Horizontally, the equilibrium condition gives:

\(T_A \times \frac{3}{5} = T_B \times \frac{4}{5}\)

Solving these equations simultaneously, we find:

\(T_A = 1.6 \text{ N}\)

\(T_B = 1.2 \text{ N}\)

To solve for the tension \(T\) and the mass \(m\), we resolve forces horizontally and vertically.

Horizontally, the equilibrium condition gives:

\(T \sin 30^{\circ} + T \sin 40^{\circ} - 2 = 0\)

Vertically, the equilibrium condition gives:

\(T \cos 30^{\circ} - T \cos 40^{\circ} - mg = 0\)

Solving the horizontal equation for \(T\):

\(T (\sin 30^{\circ} + \sin 40^{\circ}) = 2\)

\(T = \frac{2}{\sin 30^{\circ} + \sin 40^{\circ}}\)

Substituting the values of \(\sin 30^{\circ} = 0.5\) and \(\sin 40^{\circ} \approx 0.6428\):

\(T = \frac{2}{0.5 + 0.6428} \approx 1.75\)

Substitute \(T = 1.75\) into the vertical equation:

\(1.75 \cos 30^{\circ} - 1.75 \cos 40^{\circ} = mg\)

\(mg = 1.75 (\cos 30^{\circ} - \cos 40^{\circ})\)

Using \(\cos 30^{\circ} \approx 0.866\) and \(\cos 40^{\circ} \approx 0.766\):

\(mg = 1.75 (0.866 - 0.766)\)

\(mg = 1.75 \times 0.1 = 0.175\)

\(m = \frac{0.175}{g}\)

Assuming \(g = 9.8\):

\(m \approx \frac{0.175}{9.8} \approx 0.0175\)

To solve for the tension in the string and the value of X, we resolve the forces acting on the ring R.

1. Resolve vertically: \(X \sin 60^{\circ} + T \sin 30^{\circ} - 0.2g = 0\)

2. Resolve horizontally: \(X \cos 60^{\circ} - T - T \cos 30^{\circ} = 0\)

Solving these equations simultaneously:

From the vertical resolution, substitute the known values: \(X \cdot \frac{\sqrt{3}}{2} + T \cdot \frac{1}{2} = 0.2 \cdot 9.8\)

From the horizontal resolution: \(X \cdot \frac{1}{2} = T + T \cdot \frac{\sqrt{3}}{2}\)

Solving these equations gives: \(X = 2\) \(T = 0.536 \text{ N}\)

To solve this problem, we need to resolve the forces acting on the ring R both vertically and horizontally.

Vertically: The vertical components of the tension must balance the weight of the ring. The weight of the ring is given by:

\(0.2g\)

where \(g\) is the acceleration due to gravity (approximately 9.8 m/s²). The vertical components of the tension are:

\(T \sin 70^{\circ} + T \sin 45^{\circ} = 0.2g\)

Solving for \(T\):

\(T(\sin 70^{\circ} + \sin 45^{\circ}) = 0.2 \times 9.8\)

\(T = \frac{0.2 \times 9.8}{\sin 70^{\circ} + \sin 45^{\circ}} \approx 1.21 \text{ N}\)

Horizontally: The horizontal components of the tension must balance the horizontal force \(P\):

\(P + T \cos 70^{\circ} = T \cos 45^{\circ}\)

Substituting the value of \(T\):

\(P + 1.21 \cos 70^{\circ} = 1.21 \cos 45^{\circ}\)

Solving for \(P\):

\(P = 1.21 \cos 45^{\circ} - 1.21 \cos 70^{\circ} \approx 0.443 \text{ N}\)