Galileo writes that for angles of projection of a projectile at angles `(45 + theta)` and `(45 - theta)`, the horizontal ranges described by the projectile are in the ratio of (if `theta le 45`)

Options

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

Explanation

To solve the problem, we need to find the ratio of the horizontal ranges of a projectile launched at angles \( (45 + \theta) \) and \( (45 - \theta) \).

### Step-by-Step Solution:

1. **Identify the Angles of Projection**:
- Let the first angle of projection be \( \theta_1 = 45 + \theta \).
- Let the second angle of projection be \( \theta_2 = 45 - \theta \).

2. **Use the Range Formula**:
- The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( u \) is the initial velocity, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of projection.

3. **Calculate the Range for the First Angle**:
- For \( \theta_1 = 45 + \theta \):
\[ R_1 = \frac{u^2 \sin(2(45 + \theta))}{g} \]
- Simplifying \( \sin(2(45 + \theta)) \):
\[ \sin(2(45 + \theta)) = \sin(90 + 2\theta) = \cos(2\theta) \]
- Therefore, the range \( R_1 \) becomes:
\[ R_1 = \frac{u^2 \cos(2\theta)}{g} \]

4. **Calculate the Range for the Second Angle**:
- For \( \theta_2 = 45 - \theta \):
\[ R_2 = \frac{u^2 \sin(2(45 - \theta))}{g} \]
- Simplifying \( \sin(2(45 - \theta)) \):
\[ \sin(2(45 - \theta)) = \sin(90 - 2\theta) = \cos(2\theta) \]
- Therefore, the range \( R_2 \) becomes:
\[ R_2 = \frac{u^2 \cos(2\theta)}{g} \]

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

6. **Conclusion**:
- The ratio of the horizontal ranges \( R_1 \) and \( R_2 \) is \( 1:1 \).

### Final Answer:
The ratio of the horizontal ranges described by the projectile at angles \( (45 + \theta) \) and \( (45 - \theta) \) is \( 1:1 \).