To determine the angle between the two vectors given that their magnitudes are equal and the magnitude of their resultant is equal to the magnitude of either vector, we start using the vector addition formula.

Let's denote the magnitude of each vector as A. According to the problem, the resultant vector's magnitude is also A. The formula for the magnitude of the resultant vector of two vectors A and B with an angle \theta between them is given by:

R = \sqrt{A^2 + B^2 + 2AB \cos \theta}

Since the vectors have equal magnitudes, B = A. In this case, the equation becomes:

R = \sqrt{A^2 + A^2 + 2A \cdot A \cdot \cos \theta}

Simplifying, we get:

R = \sqrt{2A^2 + 2A^2 \cos \theta}

R = \sqrt{2A^2(1 + \cos \theta)}

Given that the resultant magnitude is also A, we set R = A, and equate:

A = \sqrt{2A^2(1 + \cos \theta)}

Squaring both sides, we have:

A^2 = 2A^2(1 + \cos \theta)

Dividing both sides by A^2 gives:

1 = 2(1 + \cos \theta)

Simplifying further:

1 = 2 + 2\cos \theta

1 - 2 = 2 \cos \theta

-1 = 2 \cos \theta

\cos \theta = -\frac{1}{2}

The angle whose cosine is -\frac{1}{2} is 120^\circ. Therefore, the angle between the two vectors is 120 degrees.