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

`R^(2) = A^(2) + B^(2) + 2AB cos theta`
`A = (x+ y), B= (x- y), R= sqrt(x^(2) + y^(2))`
`x^(2) + y^(2) = (x+ y)^(2) + (x- y)^(2)+ 2(x+ y) (x- y) cos theta`
`rArr x^(2) + y^(2) = 2x^(2) + 2y^(2) + 2(x^(2)- y^(2)) cos theta`
`cos theta= (- (x^(2) + y^(2)))/(2(x^(2)- y^(2))) rArr theta= cos^(-1) [(-(x^(2) + y^(2)))/(2(x^(2)- y^(2)))]`