Correct Answer: Option A

Explanation

Given forces:
\(F_1 = 10 \, \text{N} \, (\theta_1 = 20^\circ), \quad F_2 = 7 \, \text{N} \, (\theta_2 = 200^\circ)\)

Components:

1. For \( F_1 \):
\(F_{1x} = 10 \cos(20^\circ) \approx 9.40 \, \text{N}, \quad F_{1y} = 10 \sin(20^\circ) \approx 3.42 \, \text{N}\)

2. For \( F_2 \):

\(F_{2x} = 7 \cos(200^\circ) \approx -6.58 \, \text{N}, \quad F_{2y} = 7 \sin(200^\circ) \approx -2.39 \, \text{N}\)

Resultant Components:
\(R_x = F_{1x} + F_{2x} = 9.40 - 6.58 \approx 2.82 \, \text{N}\)
\(R_y = F_{1y} + F_{2y} = 3.42 - 2.39 \approx 1.03 \, \text{N}\)

Magnitude:
\(R = \sqrt{R_x^2 + R_y^2} = \sqrt{(2.82)^2 + (1.03)^2} \approx 3.00 \, \text{N}\)

Direction:
\(\theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) = \tan^{-1}\left(\frac{1.03}{2.82}\right) \approx 20.1^\circ\)
\(R \approx 3.00 \, \text{N} \, \text{at} \, 20.1^\circ\)

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