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At what angle must the two forces $(x + y)$ and $(x -y)$ act so that the resultant may be $\sqrt {({x^2} + {y^2})} $
${\cos ^{ - 1}}\left( { - \frac{{{x^2} + {y^2}}}{{2({x^2} - {y^2})}}} \right)$
${\cos ^{ - 1}}\left( { - \frac{{2({x^2} - {y^2})}}{{{x^2} + {y^2}}}} \right)$
${\cos ^{ - 1}}\left( { - \frac{{{x^2} + {y^2}}}{{{x^2} - {y^2}}}} \right)$
${\cos ^{ - 1}}\left( { - \frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}} \right)$
Solution
$ |\mathop { {\rm{R}}}\limits^ \to | = \sqrt {{A^2} + \; {B^2} + \; 2AB\cos \theta }$
$\sqrt {{x^2} + {y^2}} = \sqrt {{{\left( {x + y} \right)}^2} + \; {{\left( {x – y} \right)}^2} + \; 2 \left( {x + \; y} \right)\left( {x – y} \right)\cos \theta }$
$ \Rightarrow {x^2} + \; {y^2} = 2{x^2} + \; 2{y^2} + \; 2\left( {{x^2} – {y^2}} \right) \cos \theta$
$ \Rightarrow – \left( {{x^2} + \; {y^2}} \right) = 2\left( {{x^2} – {y^2}} \right) \cos \theta$
$ \Rightarrow \cos \theta = \frac{{ – \left( {{x^2} + \; {y^2}} \right)}}{{2 \left( {{x^2} – {y^2}} \right)}}$
$\Rightarrow \theta = {\cos ^{ – 1}} \left( {\frac{{ – \left( {{x^2} + {y^2}} \right)}}{{2 \left( {{x^2} – {y^2}} \right)}}} \right)$
