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

Answer 1

Answer: (A)

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