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

Explanation

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 will follow these steps:

### Step 1: Write the formula for range
The horizontal 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.

### Step 2: Define the angles of projection
Let:
- \( \theta_1 = 45^\circ - \theta \)
- \( \theta_2 = 45^\circ + \theta \)

### Step 3: Calculate the range for \( \theta_1 \)
Using the range formula for \( \theta_1 \):
\[ R_1 = \frac{u^2 \sin(2(45^\circ - \theta))}{g} \]
This simplifies to:
\[ R_1 = \frac{u^2 \sin(90^\circ - 2\theta)}{g} \]
Using the identity \( \sin(90^\circ - x) = \cos x \):
\[ R_1 = \frac{u^2 \cos(2\theta)}{g} \]

### Step 4: Calculate the range for \( \theta_2 \)
Now, for \( \theta_2 \):
\[ R_2 = \frac{u^2 \sin(2(45^\circ + \theta))}{g} \]
This simplifies to:
\[ R_2 = \frac{u^2 \sin(90^\circ + 2\theta)}{g} \]
Using the identity \( \sin(90^\circ + x) = \cos x \):
\[ R_2 = \frac{u^2 \cos(2\theta)}{g} \]

### Step 5: Find the ratio of the ranges
Now, we can find the ratio of the two ranges:
\[ \frac{R_1}{R_2} = \frac{\frac{u^2 \cos(2\theta)}{g}}{\frac{u^2 \cos(2\theta)}{g}} = 1 \]

### Conclusion
Thus, the ratio of the horizontal ranges \( R_1 \) and \( R_2 \) is:
\[ R_1 : R_2 = 1 : 1 \]

### Final Answer
The correct option is \( 1 : 1 \).

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