JCECE Medical JCECE Medical Solved Paper-2010

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

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

    B)  d\[{{\cos }^{-1}}\left[ \frac{-2({{x}^{2}}-{{y}^{2}})}{{{x}^{2}}+{{y}^{2}}} \right]\]

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

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

    Correct Answer: A

    Solution :

     \[{{R}^{2}}={{A}^{2}}+{{B}^{2}}+2AB\cos \theta \] Substituting,\[A=(x+y),B=(x-y)\]and \[R=\sqrt{({{x}^{2}}+{{y}^{2}})}\] we get,\[\theta ={{\cos }^{-1}}\left[ -\frac{({{x}^{2}}+{{y}^{2}})}{2({{x}^{2}}-{{y}^{2}})} \right]\]


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