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

We know that,
`R^(2)=A^(2)+B^(2)+2ABcostheta` . . (i)
On substituting the value
`A=(x+y),B=(x-y)`
and `R=sqrt((x^(2)+y^(2)))` in Eq.
(i), we get
`x^(2)+y^(2)=(x+y)^(2)+(x-y)^(2)+2(x+y)`
`(x-y)costheta`
`impliesx^(2)+y^(2)=x^(2)+y^92)+2xy+x^(2)+y^(2)-2xy+2`
`(x^(2)-y^(2))costheta`
`=(-(x^(2)+y^(2)))/(2(x^(2)-y^(2)))=costheta`
We get `theta=cos^(-1)[-((x^(2)+y^(2)))/(2(x^(2)-y^(2)))]`