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