For angles of projection of a projectile at angles `(45^(@)` – θ) and `(45^(@) + θ)`, the horizontal ranges described by the projectile are in the ratio of:

Options

A. ` 1 : 1 `
B. `2 : 3 `
C. ` 1 : 2 `
D. `2 : 1 `

Explanation

To solve the problem of finding the ratio of the horizontal ranges of a projectile launched at angles \( (45^\circ - \theta) \) and \( (45^\circ + \theta) \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Range Formula**:
The horizontal range \( R \) of a projectile launched with an initial speed \( u \) at an angle \( \alpha \) is given by the formula:
\[ R = \frac{u^2 \sin(2\alpha)}{g} \]
where \( g \) is the acceleration due to gravity.

2. **Calculate Range for \( 45^\circ - \theta \)**:
- Let \( \alpha_1 = 45^\circ - \theta \).
- Substitute into the range formula:
\[ R_1 = \frac{u^2 \sin(2(45^\circ - \theta))}{g} \]
- Simplifying \( \sin(2(45^\circ - \theta)) \):
\[ \sin(2(45^\circ - \theta)) = \sin(90^\circ - 2\theta) = \cos(2\theta) \]
- Therefore, the range \( R_1 \) becomes:
\[ R_1 = \frac{u^2 \cos(2\theta)}{g} \]

3. **Calculate Range for \( 45^\circ + \theta \)**:
- Let \( \alpha_2 = 45^\circ + \theta \).
- Substitute into the range formula:
\[ R_2 = \frac{u^2 \sin(2(45^\circ + \theta))}{g} \]
- Simplifying \( \sin(2(45^\circ + \theta)) \):
\[ \sin(2(45^\circ + \theta)) = \sin(90^\circ + 2\theta) = \cos(2\theta) \]
- Therefore, the range \( R_2 \) becomes:
\[ R_2 = \frac{u^2 \cos(2\theta)}{g} \]

4. **Find the Ratio of the Ranges**:
- Now we find the ratio \( \frac{R_1}{R_2} \):
\[ \frac{R_1}{R_2} = \frac{\frac{u^2 \cos(2\theta)}{g}}{\frac{u^2 \cos(2\theta)}{g}} = 1 \]

5. **Final Result**:
- The ratio of the horizontal ranges \( R_1 : R_2 \) is:
\[ R_1 : R_2 = 1 : 1 \]