At what angle must the two forces `(x+y)` and `(x-y)` act so that the resultant may be `sqrt((x^(2)+y^(2)))` :-

Options

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))]`

Answer

A

Explanation

`(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)))`