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