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

1

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