At what angle must the two forces (x+y) and (x-y) act so that the resultant may be `sqrt(x^(2)+y^(2))` ?

Options

A. `cos ^(-1) [-(x^(2)+y^(2))/(2(x^(2)-y^(2)))]`
B. `cos ^(-1) [-(-2x^(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)))]`

Explanation

To solve the problem, we need to find the angle at which two forces, represented as vectors \( \vec{A} = x + y \) and \( \vec{B} = x - y \), must act so that their resultant is \( R = \sqrt{x^2 + y^2} \).

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Let \( \vec{A} = x + y \)
- Let \( \vec{B} = x - y \)
- The resultant \( R = \sqrt{x^2 + y^2} \)

2. **Use the Resultant Formula**:
The formula for the resultant \( R \) of two vectors \( \vec{A} \) and \( \vec{B} \) when they make an angle \( \theta \) with each other is given by:
\[ R = \sqrt{A^2 + B^2 + 2AB \cos \theta} \]

3. **Substitute the Values**:
Substitute \( A = x + y \) and \( B = x - y \):
\[ R = \sqrt{(x+y)^2 + (x-y)^2 + 2(x+y)(x-y) \cos \theta} \]

4. **Calculate \( A^2 \) and \( B^2 \)**:
- \( A^2 = (x+y)^2 = x^2 + 2xy + y^2 \)
- \( B^2 = (x-y)^2 = x^2 - 2xy + y^2 \)

5. **Combine \( A^2 \) and \( B^2 \)**:
\[ A^2 + B^2 = (x^2 + 2xy + y^2) + (x^2 - 2xy + y^2) = 2x^2 + 2y^2 \]

6. **Calculate \( 2AB \cos \theta \)**:
\[ 2AB = 2(x+y)(x-y) = 2(x^2 - y^2) \]
Thus, the equation for \( R \) becomes:
\[ R = \sqrt{2(x^2 + y^2) + 2(x^2 - y^2) \cos \theta} \]

7. **Set the Resultant Equal to \( \sqrt{x^2 + y^2} \)**:
\[ \sqrt{x^2 + y^2} = \sqrt{2(x^2 + y^2) + 2(x^2 - y^2) \cos \theta} \]

8. **Square Both Sides**:
\[ x^2 + y^2 = 2(x^2 + y^2) + 2(x^2 - y^2) \cos \theta \]

9. **Rearranging the Equation**:
\[ x^2 + y^2 - 2(x^2 + y^2) = 2(x^2 - y^2) \cos \theta \]
\[ -x^2 - y^2 = 2(x^2 - y^2) \cos \theta \]

10. **Solve for \( \cos \theta \)**:
\[ \cos \theta = -\frac{x^2 + y^2}{2(x^2 - y^2)} \]

11. **Find the Angle \( \theta \)**:
\[ \theta = \cos^{-1}\left(-\frac{x^2 + y^2}{2(x^2 - y^2)}\right) \]

### Final Answer:
The angle \( \theta \) at which the two forces must act is given by:
\[ \theta = \cos^{-1}\left(-\frac{x^2 + y^2}{2(x^2 - y^2)}\right) \]