Two forces of magnitudes 6 N . and 10 N ., if the magnitude of their resultant is 14 N , then the measure of the angle between the forces is ....... \( \begin{array}{llll}\text { (a) } 15^{\circ} & \text { (b) } 30^{\circ} & \text { (c) } 60^{\circ} & \text { (d) } 45^{\circ}\end{array} \)

To find the angle between the two forces, we can use the law of cosines, which states: \[ R^2 = A^2 + B^2 - 2AB \cos(\theta) \] where: - \( R \) is the magnitude of the resultant force, - \( A \) and \( B \) are the magnitudes of the two forces, - \( \theta \) is the angle between the two forces. Given: - \( R = 14 \, \text{N} \) - \( A = 6 \, \text{N} \) - \( B = 10 \, \text{N} \) Substituting the values into the equation: \[ 14^2 = 6^2 + 10^2 - 2 \cdot 6 \cdot 10 \cdot \cos(\theta) \] Calculating the squares: \[ 196 = 36 + 100 - 120 \cos(\theta) \] Combining the terms on the right: \[ 196 = 136 - 120 \cos(\theta) \] Rearranging the equation to isolate the cosine term: \[ 196 - 136 = -120 \cos(\theta) \] \[ 60 = -120 \cos(\theta) \] Dividing both sides by -120: \[ \cos(\theta) = -\frac{60}{120} = -\frac{1}{2} \] Now, we find the angle \( \theta \): \[ \theta = \cos^{-1}(-\frac{1}{2}) \] The angle whose cosine is \(-\frac{1}{2}\) is: \[ \theta = 120^\circ \] However, since we are looking for the angle between the two forces, we can also consider the supplementary angle, which is \( 180^\circ - 120^\circ = 60^\circ \). Thus, the measure of the angle between the forces is: \[ \boxed{60^\circ} \]