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

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Question

$ |\mathop { {m{R}}}\limits^ 	o | = \sqrt {{A^2} + \; {B^2} + \; 2AB\cos 	heta }$

$\sqrt {{x^2} + {y^2}} = \sqrt {{{\left( {x + y} ight)

Vedclass · Accepted Answer

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

Vedclass · Answer

$ |\mathop { {m{R}}}\limits^ 	o | = \sqrt {{A^2} + \; {B^2} + \; 2AB\cos 	heta }$

$\sqrt {{x^2} + {y^2}} = \sqrt {{{\left( {x + y} ight)}^2} + \; {{\left( {x - y} ight)}^2} + \; 2 \left( {x + \; y} ight)\left( {x - y} ight)\cos 	heta }$

$ \Rightarrow {x^2} + \; {y^2} = 2{x^2} + \; 2{y^2} + \; 2\left( {{x^2} - {y^2}} ight) \cos 	heta$

$ \Rightarrow - \left( {{x^2} + \; {y^2}} ight) = 2\left( {{x^2} - {y^2}} ight) \cos 	heta$

$ \Rightarrow \cos 	heta = \frac{{ - \left( {{x^2} + \; {y^2}} ight)}}{{2 \left( {{x^2} - {y^2}} ight)}}$

$\Rightarrow 	heta = {\cos ^{ - 1}} \left( {\frac{{ - \left( {{x^2} +  {y^2}} ight)}}{{2 \left( {{x^2} - {y^2}} ight)}}} ight)$

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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