Let the two forces have equal magnitude A. If | vedclass

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(B) The magnitude of the resultant $R$ of two vectors of equal magnitude $A$ with an angle $	heta$ between them is given by the formula: $R = \sqrt{A

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$\cos^{-1}\left(-\frac{7}{9}\right)$

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(B) The magnitude of the resultant $R$ of two vectors of equal magnitude $A$ with an angle $	heta$ between them is given by the formula: $R = \sqrt{A^2 + A^2 + 2A^2 \cos 	heta}$.Given $R = \frac{2A}{3}$,we substitute this into the equation:$\frac{2A}{3} = \sqrt{2A^2 + 2A^2 \cos 	heta}$.Squaring both sides:$\frac{4A^2}{9} = 2A^2(1 + \cos 	heta)$.Dividing both sides by $2A^2$:$\frac{2}{9} = 1 + \cos 	heta$.Solving for $\cos 	heta$:$\cos 	heta = \frac{2}{9} - 1 = \frac{2-9}{9} = -\frac{7}{9}$.Therefore,the angle $	heta = \cos^{-1}\left(-\frac{7}{9}ight)$.

Let the two forces have equal magnitude $A$. If the magnitude of the resultant is $\frac{2A}{3}$,then the angle between those two forces is:

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