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)[-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

To solve the problem, we need to find the angle at which the two forces \( (x+y) \) and \( (x-y) \) must act so that their resultant is equal to \( \sqrt{x^2 + y^2} \).

### Step-by-Step Solution:

1. **Identify the Forces and Resultant**:
- Let \( F_1 = x + y \) (magnitude of the first force)
- Let \( F_2 = x - y \) (magnitude of the second force)
- The resultant \( R \) is given as \( R = \sqrt{x^2 + y^2} \).

2. **Use the Formula for Resultant of Two Forces**:
The resultant \( R \) of two forces can be calculated using the formula:
\[ R^2 = F_1^2 + F_2^2 + 2 F_1 F_2 \cos \theta \]
where \( \theta \) is the angle between the two forces.

3. **Substitute the Values**:
Substitute \( F_1 \) and \( F_2 \) into the equation:
\[ R^2 = (x+y)^2 + (x-y)^2 + 2(x+y)(x-y) \cos \theta \]

4. **Expand the Squares**:
Expanding \( (x+y)^2 \) and \( (x-y)^2 \):
\[ (x+y)^2 = x^2 + 2xy + y^2 \]
\[ (x-y)^2 = x^2 - 2xy + y^2 \]
Therefore,
\[ R^2 = (x^2 + 2xy + y^2) + (x^2 - 2xy + y^2) + 2(x+y)(x-y) \cos \theta \]

5. **Combine Like Terms**:
Combine the terms:
\[ R^2 = 2x^2 + 2y^2 + 2(x+y)(x-y) \cos \theta \]

6. **Calculate \( (x+y)(x-y) \)**:
\[ (x+y)(x-y) = x^2 - y^2 \]
Thus, we can rewrite the equation as:
\[ R^2 = 2x^2 + 2y^2 + 2(x^2 - y^2) \cos \theta \]

7. **Set Up the Equation**:
Since \( R = \sqrt{x^2 + y^2} \), squaring both sides gives:
\[ x^2 + y^2 = 2x^2 + 2y^2 + 2(x^2 - y^2) \cos \theta \]

8. **Rearranging the Equation**:
Rearranging gives:
\[ 0 = x^2 + y^2 + 2(x^2 - y^2) \cos \theta \]

9. **Isolate \( \cos \theta \)**:
\[ 2(x^2 - y^2) \cos \theta = - (x^2 + y^2) \]
Thus,
\[ \cos \theta = - \frac{x^2 + y^2}{2(x^2 - y^2)} \]

10. **Find the Angle \( \theta \)**:
Therefore, the angle \( \theta \) is given by:
\[ \theta = \cos^{-1} \left(- \frac{x^2 + y^2}{2(x^2 - y^2)}\right) \]