The resultant of two forces, one double the other in magnitude, is perpendicular to the smaller of the two forces. The angle between the two forces is

Let the two forces be represented as $A$ and $B$, where $A$ is the smaller force and $B$ is the larger force.
According to the problem, the magnitude of the larger force is double the magnitude of the smaller force. Therefore, we can write:
$B=2A$

Let $\theta$ be the angle between the two forces $A$ and $B$.
The direction of the resultant force $R$ making an angle $\alpha$ with the smaller force $A$ is given by the formula:
$\tan \left(\alpha \right)=\frac{B\sin \left(\theta \right)}{A+B\cos \left(\theta \right)}$

We are given that the resultant is perpendicular to the smaller force $A$, which means:
$\alpha ={90}^{\circ }$

Since $\tan \left({90}^{\circ }\right)$ is undefined (approaching infinity), the denominator of the expression must be equal to zero:
$A+B\cos \left(\theta \right)=0$

Substitute $B=2A$ into the equation:
$A+2A\cos \left(\theta \right)=0$

Divide the entire equation by $A$ (since $A\ne 0$):
$1+2\cos \left(\theta \right)=0$

Solve for $\cos \left(\theta \right)$:
$2\cos \left(\theta \right)=-1$
$\cos \left(\theta \right)=-\frac{1}{2}$

The angle $\theta$ whose cosine is $-\frac{1}{2}$ in the range of $0^{\circ }$ to ${180}^{\circ }$ is:
$\theta ={120}^{\circ }$

Thus, the angle between the two forces is 120°.