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 \(T_A\):

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