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

We know that  ${{R}^{2}}={{A}^{2}}+{{B}^{2}}+2AB\cos \theta$ On substituting the value $A=(x+y),\,B=(x-y)$ ${{x}^{2}}+{{y}^{2}}\,={{(x+y)}^{2}}\,+{{(x-y)}^{2}}+2(x-y)\,\cos \theta$ $\Rightarrow \,{{x}^{2}}+{{y}^{2}}\,={{x}^{2}}+{{y}^{2}}+2xy\,+{{x}^{2}}+{{y}^{2}}\,-2xy$ $+2({{x}^{2}}-{{y}^{2}})\,\cos \theta$ $\Rightarrow \,\,\frac{-({{x}^{2}}+{{y}^{2}})}{2({{x}^{2}}-{{y}^{2}})}\,=\cos \theta$ $\Rightarrow \,\theta =\,{{\cos }^{-1}}\,\left[ -\,\frac{({{x}^{2}}+{{y}^{2}})}{2({{x}^{2}}-{{y}^{2}})} \right]$