Physics, asked by Sreeunni6144, 11 months ago

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})}]

Answers

Answered by aayushg1713
72

The correct option is (a)

Mark this answer as BRAINLIEST please !!!

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misha08: ty
Answered by CarliReifsteck
30

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)}

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