At what angle do two forces (x + y) and (x - | vedclass

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(A) The resultant $R$ of two vectors $A$ and $B$ acting at an angle $	heta$ is given by $R = \sqrt{A^2 + B^2 + 2AB \cos 	heta}$.Given $A = x + y$,$B

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$\cos^{-1} \left( \frac{-(x^2 + y^2)}{2(x^2 - y^2)} \right)$

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(A) The resultant $R$ of two vectors $A$ and $B$ acting at an angle $	heta$ is given by $R = \sqrt{A^2 + B^2 + 2AB \cos 	heta}$.Given $A = x + y$,$B = x - y$,and $R = \sqrt{x^2 + y^2}$.Squaring both sides: $R^2 = A^2 + B^2 + 2AB \cos 	heta$.Substitute the values: $x^2 + y^2 = (x + y)^2 + (x - y)^2 + 2(x + y)(x - y) \cos 	heta$.Expand the terms: $x^2 + y^2 = (x^2 + 2xy + y^2) + (x^2 - 2xy + y^2) + 2(x^2 - y^2) \cos 	heta$.Simplify: $x^2 + y^2 = 2x^2 + 2y^2 + 2(x^2 - y^2) \cos 	heta$.Rearrange to solve for $\cos 	heta$: $x^2 + y^2 - 2x^2 - 2y^2 = 2(x^2 - y^2) \cos 	heta$.$-(x^2 + y^2) = 2(x^2 - y^2) \cos 	heta$.$\cos 	heta = \frac{-(x^2 + y^2)}{2(x^2 - y^2)}$.Therefore,$	heta = \cos^{-1} \left( \frac{-(x^2 + y^2)}{2(x^2 - y^2)} ight)$.

At what angle do two forces $(x + y)$ and $(x - y)$ act so that their resultant is $\sqrt{x^2 + y^2}$?

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