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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)))` :-A. `"cos"^(-1)(-(x^(2)+y^(2)))/(2(x^(2)-y^(2)))`B. `"cos"^(-1)(-2(x^(2)-y^(2)))/(x^(2)+y^(2))`C. `"cos"^(-1)(-(x^(2)+y^(2)))/(x^(2)-y^(2))`D. `"cos"^(-1)((x^(2)-y^(2)))/(x^(2)+y^(2))` |
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Answer» Correct Answer - A Resultant `=sqrt((x^(2)+y^(2)))` `=sqrt((x+y)^(2)+(x-y)^(2)+2(x+y)(x-y) cos theta)` `rArr x^(2)+y^(2)=2(x^(2)+y^(2))+2(x^(2)-y^(2))cos theta` `rArrcos theta=1/2((x^(2)+y^(2))/(y^(2)-x^(2)))` |
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