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. $\cos^{-1} \Bigg[ - \frac{x^2 + y^2}{2(x^2 - y^2)}\Bigg]$
• B. $\cos^{-1} \Bigg[ - \frac{2(x^2 - y^2)}{x^2 + y^2}\Bigg]$
• C. $\cos^{-1} \Bigg[ - \frac{(x^2 + y^2)}{(x^2 - y^2)}\Bigg]$
• D. $cos^{-1} \Bigg[ - \frac{(x^2 - y^2)}{(x^2 + y^2)}\Bigg]$

Answer 1

Answer: (A)

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