At what angle must the two forces (x + y) and (x - y) act so that the resultant may be `(x^(2) + y^(2))^(1//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

`R^(2) =A^(@) + B^(2) + 2ABcostheta`
`implies (x^(2) + y^(2))=(x+y)^(2) + (x-y)^(2) + 2(x^(2)-y^(2)).costheta`
`x^(2) + y^(2) =2(x^(2) + y^(2)) + 2(x^(2)-y^(2)).costheta`
`implies costheta=(-(x^(2) + y^(2)))/(2(x^(2) - y^(2)))`
`implies theta=cos^(-1)((-x^(2)-y^(2))/(2(x^(2) + y^(2))))`