A particle is acted upon by two forces 6N and 3N inclined at an angle of 120° to each other. Find the magnitude of the resultant force.

To find the magnitude of the resultant force, we can use the Law of Cosines, which states that in any triangle, the square of one side is equal to the sum of the squares of the other two sides minus twice the product of those two sides and the cosine of the included angle. In this case, let the two forces be a and b, with an included angle of 120°. Then, the magnitude of the resultant force, which we'll call R, is: $$R^2 = a^2 + b^2 - 2ab\cos{120^\circ}$$ Since cosine of 120° is -1/2, we can simplify this to: $$R^2 = a^2 + b^2 + ab$$ Substituting the values given in the problem, we get: $$R^2 = (6\text{ N})^2 + (3\text{ N})^2 + (6\text{ N})(3\text{ N}) = 54\text{ N}^2$$ Taking the square root of both sides, we get: $$R = \sqrt{54\text{ N}^2} = 3\sqrt{6}\text{ N} \approx 7.746\text{ N}$$ Therefore, the correct answer is (D) \(3\sqrt{3}\) N.