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