Two forces of equal magnitude simultaneously act at a point. It is...

Let the two forces be $\vec{F_1}$ and $\vec{F_2}$, and their resultant be $\vec{R}$.

Given that $|\vec{F_1}| = |\vec{F_2}| = |\vec{R}| = F$.

The magnitude of the resultant of two vectors is given by $R^2 = F_1^2 + F_2^2 + 2F_1 F_2 \cos \theta$, where $\theta$ is the angle between the two forces.

Substituting the given values:

$F^2 = F^2 + F^2 + 2F^2 \cos \theta$

$F^2 = 2F^2(1 + \cos \theta)$

$\cos \theta = -\dfrac{1}{2}$

$\theta = \dfrac{2\pi}{3}$

Thus, the angle between the individual forces is $\dfrac{2\pi}{3}$. Statement II is correct.

Since the two forces are equal in magnitude, their resultant bisects the angle between them. Let $\alpha$ be the angle between the resultant and one of the forces.

$\alpha = \dfrac{\theta}{2} = \dfrac{1}{2} \left(\dfrac{2\pi}{3}\right) = \dfrac{\pi}{3}$

Thus, the angle between one of the individual forces and the resultant force is $\dfrac{\pi}{3}$. Statement I is also correct.

Both statements I and II are correct.