Question

the angle between the forces (x+y) and (x-y) if their resultant is sqrt{x^{2}+y^{2}}

Answer 1

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Formula used :
R² = a1² +a2² +2a1a2cos¢

Where A1 and A2 are the forces, R is the resultant force and ¢ is the angle between them.
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By problem,
${ (\sqrt{ {x}^{2} + {y}^{2} }) }^{2} = {(x + y)}^{2} + {(x - y)}^{2} + 2(x + y)(x - y) \cos( \alpha ) \\ or. \: {x}^{2} + {y}^{2} = {x}^{2} + {y}^{2} + 2xy + {x}^{2} + {y}^{2} - 2xy + 2( {x}^{2} - {y}^{2} ) \cos( \alpha ) \\ or. \: - ( {x}^{2} + {y}^{2} ) = 2( {x}^{2} - {y}^{2} ) \cos( \alpha ) \\ or. \cos( \alpha ) = - \frac{ {x}^{2} + {y}^{2} }{2( {x}^{2} - {y}^{2}) } \\or. \: \alpha = \cos^{ - 1} ( - \frac{ {x}^{2} + {y}^{2} }{2( {x}^{2} - {y}^{2}) } )$
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