Question

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

Answer 1

Answer:

θ = cos⁻¹ ((x²+y²)/2(y²-x²))

(where θ is the angle b/w Forces (x+y) and (x-y))

Explanation:

Let , the angle b/w the forces of magnitudes (x+y) and (x-y) be θ

it is given that the magnitude of resultant should be √(x²+y²)

Using formula to calculate magnitude of resultant of two vectors

R² = A² + B² + 2 AB cos θ

( where R is magnitude of resultant of two vectors of magnitudes A and B , θ is the angle b/w vectors A and B )

so,

→ (√(x²+y²) )² = (x+y)²+(x-y)²+2(x+y)(x-y) cos θ

→ x²+y² = x²+y²+2xy+x²+y²-2xy+2(x²-y²) cos θ

→ x²+y² = 2(x²+y²) + 2(x²-y²) cos θ

→ x²+y² - 2(x²+y²) = 2(x²-y²) cos θ

→ cos θ = -(x²+y²)/2(x²-y²)

→ cos θ = (x²+y²) / 2(y²-x²)

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→ θ = cos⁻¹ ((x²+y²)/2(y²-x²))

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Answer.