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

Resultant `=sqrt((x^(2)+y^(2)))`
`=sqrt((x+y)^(2)+(x-y)^(2)+2(x+y)(x-y) cos theta)`
`rArr x^(2)+y^(2)=2(x^(2)+y^(2))+2(x^(2)-y^(2))cos theta`
`rArrcos theta=1/2((x^(2)+y^(2))/(y^(2)-x^(2)))`