Concept:
Solving for a component vector magnitude using the law of cosines.

Formula:
$$B^2 = A^2 + R^2 - 2AR \cos\phi$$

Solution:
Let the resultant force be \(R = 20\text{ N}\) and the known force be \(A = 20\sqrt{3}\text{ N}\). The angle between them is \(\phi = 30^\circ\).
Using the law of cosines on the vector triangle:
$$B^2 = (20\sqrt{3})^2 + (20)^2 - 2(20\sqrt{3})(20) \cos(30^\circ)$$
$$B^2 = (400 \times 3) + 400 - 800\sqrt{3} \left(\frac{\sqrt{3}}{2}\right)$$
$$B^2 = 1200 + 400 - 1200$$
$$B^2 = 400 \implies B = 20\text{ N}$$

Why other options are incorrect:

• A, C, and D are mathematically incorrect and do not satisfy the law of cosines for this vector geometry.