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કયા ખૂણે બે બળો $(x + y)$ અને $(x - y)$ એ પ્રક્રિયા કરે છે. તેથી તેમનું પરિણામી લગભગ $\sqrt {\left( {{x^2}\,\, + \;\,{y^2}} \right)} $ મળે ?
${\cos ^{ - 1}}\,\,\frac{{ - \left( {{x^2}\,\, + \;\,{y^2}\,\,} \right)}}{{2\,\,\left( {{x^2}\,\, - \,\,{y^2}} \right)}}$
${\cos ^{ - 1}}\frac{{ - 2\,\,\left( {{x^2}\,\, - \,\,{y^2}} \right)}}{{{x^2}\,\, + \;\,{y^2}}}$
${\cos ^{ - 1}}\,\,\frac{{ - \left( {{x^2}\, + \,{y^2}} \right)}}{{{x^2}\,\, - \,\,{y^2}}}$
${\cos ^{ - 1}}\frac{{\left( {{x^2}\,\, - \,\,{y^2}} \right)}}{{{x^2}\,\, + \;\,{y^2}}}$
Solution
$|\mathop {\,{\text{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) $
