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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 `(sqrt((x^(2)+y^(2))))^(2)=` `(sqrt((x+y)^(2)+(X-y)^(2)+2(x+y)(x-y)costheta))^(2)` `x^(2)+y^(2)=2(x^(2)y^(2))+2(x^(2)-y^(2))cos theta` `cos theta=-((x^(2)+y^(2)))/(2(x^(2)-y^(2)))` |
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