8. Two equal forces, the magnitude of each of them is 6 N , the magnitude of their resultant is 6 N , then the angle between them equals: \( \begin{array}{lll}\text { A) } 30^{\circ} & \text { a. } 120^{\circ} & \text { D } 150^{\circ}\end{array} \)

To find the angle between two equal forces when their resultant is given, we can use the formula for the resultant \( R \) of two forces \( F \) at an angle \( \theta \): \[ R = \sqrt{F^2 + F^2 + 2F \cdot F \cdot \cos(\theta)} \] Given that both forces have a magnitude of \( F = 6 \, \text{N} \) and the resultant \( R = 6 \, \text{N} \), we can substitute these values into the equation: \[ 6 = \sqrt{6^2 + 6^2 + 2 \cdot 6 \cdot 6 \cdot \cos(\theta)} \] This simplifies to: \[ 6 = \sqrt{36 + 36 + 72 \cos(\theta)} \] \[ 6 = \sqrt{72 + 72 \cos(\theta)} \] Now, squaring both sides: \[ 36 = 72 + 72 \cos(\theta) \] Rearranging gives: \[ 72 \cos(\theta) = 36 - 72 \] \[ 72 \cos(\theta) = -36 \] \[ \cos(\theta) = -\frac{36}{72} = -\frac{1}{2} \] The angle \( \theta \) for which \( \cos(\theta) = -\frac{1}{2} \) is: \[ \theta = 120^\circ \quad \text{or} \quad \theta = 240^\circ \] Since we are looking for the angle between the two forces, we take \( \theta = 120^\circ \). Thus, the angle between the two forces is: \[ \boxed{120^\circ} \]