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))/(2x^(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^(2)+B^(2)+2AB cos theta`
Substituting, `A= (x+y), B= (x+y)` and
`R=sqrt((x^(2)+y^(2)))`, we get
`theta = cos^(-1)(-((x^(2)+y^(2)))/(2(x^(2)-y^(2))))`