Two forces with equal magnitudes F act on a b | vedclass

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Question

(A) The magnitude of the resultant force $R$ of two vectors $F_1$ and $F_2$ with an angle $	heta$ between them is given by the formula: $R^2 = F_1^2

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

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(A) The magnitude of the resultant force $R$ of two vectors $F_1$ and $F_2$ with an angle $	heta$ between them is given by the formula: $R^2 = F_1^2 + F_2^2 + 2F_1F_2\cos	heta$. Given that $F_1 = F_2 = F$ and $R = F/3$,we substitute these values into the equation: $(F/3)^2 = F^2 + F^2 + 2(F)(F)\cos	heta$. $F^2/9 = 2F^2 + 2F^2\cos	heta$. Dividing both sides by $F^2$: $1/9 = 2 + 2\cos	heta$. $1/9 - 2 = 2\cos	heta$. $-17/9 = 2\cos	heta$. $\cos	heta = -17/18$. Therefore,$	heta = \cos^{-1}(-17/18)$.

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

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