For angles of projection of a projectile at angle `(45^(@) - theta) and (45^(@)+ theta)`, 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`

Answer

To solve the problem of finding the ratio of horizontal ranges for 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 velocity \( u \) at an angle \( \theta \) is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( g \) is the acceleration due to gravity.

2. **Calculate the Ranges for Both Angles**:
- For the angle \( (45^\circ - \theta) \):
\[ R_1 = \frac{u^2 \sin(2(45^\circ - \theta))}{g} = \frac{u^2 \sin(90^\circ - 2\theta)}{g} = \frac{u^2 \cos(2\theta)}{g} \]
- For the angle \( (45^\circ + \theta) \):
\[ R_2 = \frac{u^2 \sin(2(45^\circ + \theta))}{g} = \frac{u^2 \sin(90^\circ + 2\theta)}{g} = \frac{u^2 \cos(2\theta)}{g} \]

3. **Form the Ratio of the Ranges**:
Now we can form the ratio of the two ranges:
\[ \frac{R_1}{R_2} = \frac{\frac{u^2 \cos(2\theta)}{g}}{\frac{u^2 \sin(2\theta)}{g}} = \frac{\cos(2\theta)}{\sin(2\theta)} = \cot(2\theta) \]

4. **Evaluate the Ratio**:
Since \( \cot(2\theta) \) does not simplify to a numerical value without specific values for \( \theta \), we can conclude that the ranges are dependent on the angle \( \theta \).

5. **Conclusion**:
The ratio of the horizontal ranges for the angles \( (45^\circ - \theta) \) and \( (45^\circ + \theta) \) is given by:
\[ \frac{R_1}{R_2} = \cot(2\theta) \]