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) $\cos^{-1}[-(x^{2} + y^{2})}/{2(x^{2} - y^{2})}]$(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 1

The correct option is (a)

Mark this answer as BRAINLIEST please !!!

Answer 2

Answer:

The angle is $\theta =cos^{-1}\dfrac{-(x^2+y^2)}{2(x^2-y^2)}$

a is correct option

Explanation:

Given that,

Forces

$F_{1}=(x+y)$

$F_{2}=(x-y)$

Resultant $R =\sqrt{(x^2+y^2)}$

We know that,

The resultant is

$R^2 = F_{1}^2+F_{2}^2 +2F_{1}F_{2}\cos\theta$

$(\sqrt{(x^2+y^2)})^2=(x+y)^2+(x-y)^2+2(x+y)(x-y)\cos\theta$

$(x^2+y^2)=x^2+y^2+2xy+x^2+y^2-2xy+2(x^2-y^2)\cos\theta$

$2(x^2-y^2)\cos\theta=-(x^2+y^2)$

$\theta =cos^{-1}\dfrac{-(x^2+y^2)}{2(x^2-y^2)}$

Hence, The angle is $\theta =cos^{-1}\dfrac{-(x^2+y^2)}{2(x^2-y^2)}$