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

Answer 1

hey dear ur answer is c part hope it helps u dear friend

Answer 2

Answer:

$\sf Option \ (a): cos^{-1}\Bigg(-\dfrac{\left(x^2 + y^2\right)}{2\left( x^2 - y^2 \right) \ \ }\Bigg)$

Explanation:

ATQ, The two forces acting are (x + y) & (x - y). Let us name them A & B respectively. And, we know that the resultant is $\sf \sqrt{x^2 + y^2}$ .

We are asked to find the angle formed by the vectors/forces. We know that if θ is the angle formed between two vectors, then the resultant of these vectors can be found out using the below formula.

$\longmapsto \sf R^2 = A^2 + B^2 + 2ABcos\theta$

Now, substituting the forces in the equation we get,

$\Longrightarrow \sf R^2 = A^2 + B^2 + 2ABcos\theta \\ \\ \\\Longrightarrow \sf \left(\sqrt{x^2 + y^2} \ \right)^2 = \Big(x + y\Big)^2 + \Big(x - y\Big)^2 + 2\Big(x+y\Big) \Big(x-y\Big)cos\theta$

Using the below identities we get;

(a + b)² = a² + b² + 2ab

(a - b)² = a² + b² - 2ab

(a + b) (a - b) = a² - b²

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

Cancelling x² + y² on both sides, and cancelling 2xy & -2xy on the right hand side we get,

$\Longrightarrow \ \sf 0 = x^2 + y^2 + 2cos\theta(x^2 - y^2)\\ \\ \\\Longrightarrow \ \sf -(x^2 + y^2) = 2cos\theta(x^2 - y^2)\\ \\ \\\Longrightarrow \ \sf 2cos\theta(x^2 - y^2) = -(x^2 + y^2)\\ \\ \\\Longrightarrow \ \sf 2cos\theta = \dfrac{-(x^2 + y^2)}{ \ \ (x^2 - y^2)}\\ \\ \\\Longrightarrow \ \sf cos\theta = \dfrac{-(x^2 + y^2)}{ \ 2(x^2 - y^2)}\\ \\ \\\Longrightarrow \ \sf \theta = cos^{-1} \ \Bigg(\dfrac{-(x^2 + y^2)}{ \ 2(x^2 - y^2)}\Bigg)$

Therefore, the answer is Option (a).