Two forces, with equal magnitude F, act on a body and the magnitude of the resultant force is F/3 . The angle between the two forces is

The correct option is cos⁻¹(-17/18).

To find the angle between the two forces, we can use the formula for the magnitude of the resultant of two vectors. Let the two forces be represented as vectors $\overrightarrow{A}$ and $\overrightarrow{B}$, with an angle $\theta$ between them. The magnitude of their resultant force $R$ is given by the formula:

$R^2=A^2+B^2+2ABcos\theta$

According to the problem statement:
- The magnitude of each force is $A=B=F$
- The magnitude of the resultant force is $R=\frac{F}{3}$

Substituting these values into the resultant formula, we get:

${\left(\frac{F}{3}\right)}^2=F^2+F^2+2\left(F\right)\left(F\right)cos\theta$

Simplifying the equation:

$\frac{F^2}{9}$

We can divide the entire equation by $F^2$ (assuming $F\ne 0$):

$\frac{1}{9}=2+2cos\theta$

Rearranging the terms to solve for $cos\theta$:

$2cos\theta =\frac{1}{9}-2$

$2cos\theta =\frac{1-18}{9}$

$2cos\theta =-\frac{17}{9}$

$cos\theta =-\frac{17}{18}$

Taking the inverse cosine of both sides to find the angle $\theta$:

$\theta ={cos}^{-1}\left(-\frac{17}{18}\right)$

Therefore, the angle between the two forces is ${cos}^{-1}\left(-\frac{17}{18}\right)$.