At what angle must the two forces (x + y) and | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The magnitude of 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,we get $R^2 = A^2 + B^2 + 2AB \cos 	heta$.Substituting the values: $x^2 + y^2 = (x + y)^2 + (x - y)^2 + 2(x + y)(x - y) \cos 	heta$.Expanding the terms: $x^2 + y^2 = (x^2 + 2xy + y^2) + (x^2 - 2xy + y^2) + 2(x^2 - y^2) \cos 	heta$.Simplifying: $x^2 + y^2 = 2x^2 + 2y^2 + 2(x^2 - y^2) \cos 	heta$.Rearranging: $x^2 + y^2 - 2x^2 - 2y^2 = 2(x^2 - y^2) \cos 	heta$.$- (x^2 + y^2) = 2(x^2 - y^2) \cos 	heta$.Therefore,$\cos 	heta = -\frac{x^2 + y^2}{2(x^2 - y^2)}$.Thus,$	heta = \cos^{-1}\left(-\frac{x^2 + y^2}{2(x^2 - y^2)}ight)$.

At what angle must the two forces $(x + y)$ and $(x - y)$ act so that the resultant may be $\sqrt{x^2 + y^2}$?

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