What is the angle between two vector forces of equal magnitude such that their resultant is one-third of either of the original forces?

The correct option is C ${\cos }^{-1}(-\frac{17}{18})$
Lets the vectors be the $x$ and $x$

and resultant will be $\frac{1}{3}x$

$R_{eq}=\sqrt{R_1^2+R_2^2+2R_1{R}_{2}Cos\theta }$

According to given condition:

$(\frac{1}{3}x{)}^2$ =$x^2+x^2+{2x}^{2}cos\theta$

$\frac{x^2}{9}$ = $2x^2+2x^{2}cos\theta$

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

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

hence (A) is the correct answer