To find the tensions in the strings, resolve the forces horizontally and vertically.

Resolving horizontally:

\(T_P \cos 60^{\circ} = T_R \cos 30^{\circ}\)

Resolving vertically:

\(T_P \sin 60^{\circ} = T_R \sin 30^{\circ} + 0.2g\)

Using \(g = 9.8 \text{ m/s}^2\), solve these equations simultaneously.

From the horizontal resolution:

\(T_P \cdot 0.5 = T_R \cdot \frac{\sqrt{3}}{2}\)

\(T_P = T_R \cdot \sqrt{3}\)

Substitute \(T_P = T_R \cdot \sqrt{3}\) into the vertical resolution:

\(T_R \cdot \sqrt{3} \cdot \frac{\sqrt{3}}{2} = T_R \cdot 0.5 + 0.2 \cdot 9.8\)

\(\frac{3}{2} T_R = 0.5 T_R + 1.96\)

\(T_R = 2 \text{ N}\)

Substitute \(T_R = 2 \text{ N}\) back to find \(T_P\):

\(T_P = 2 \cdot \sqrt{3} \approx 3.46 \text{ N}\)