Question

At what angle must the two forces (x+y) and (x-y) act so that the resultant may be $\sqrt{(x^2+y^2)}$ ?

• A. $co{s}^{-1}[-(x^2+y^2\left)/2\right(x^2-y^2\left)\right]$
• B. $co{s}^{-1}[-2(x^2-y^2\left)/\right(x^2+y^2\left)\right]$
• C. $co{s}^{-1}[-(x^2+y^2\left)/\right(x^2-y^2\left)\right]$
• D. $co{s}^{-1}[-(x^2-y^2\left)/\right(x^2-y^2\left)\right]$

Answer 1

The correct option is A $co{s}^{-1}[-(x^2+y^2\left)/2\right(x^2-y^2\left)\right]$

$(\sqrt{x^2+y^2}{)}^2=(x+y{)}^2+(x-y{)}^2+2(x+y\left)\right(x-y)cos\theta$
$x^2+y^2=x^2+y^2+2xy+x^2+y^2-2xy+2(x^2-y^2)cos\theta$
$2(x^2-y^2).cos\theta =-(x^2+y^2)$
$\Rightarrow \theta =co{s}^{-1}\left[\frac{-(x^2+y^2)}{2(x^2-y^2)}\right]$