A round balloon of radius 'r' subtends an angle `alpha` at the eye of the observer, while the angle of elevation of its centre is `beta`. Find the height of the centre of balloon.

Options

A. `r" cosec "((alpha)/(2))sin beta`
B. `r sin alpha" cosec "beta`
C. `(r)/(2)sin beta`
D. `r sec ((alpha)/(2))`

Answer

To find the height of the center of the balloon, we will follow these steps:

### Step 1: Understand the Geometry
We have a round balloon with radius \( r \) that subtends an angle \( \alpha \) at the observer's eye. The angle of elevation to the center of the balloon is \( \beta \). We need to find the height \( OP \) of the center of the balloon from the observer's eye level.

### Step 2: Divide the Angle
Since the angle \( \alpha \) subtended by the balloon is at the observer's eye, we can divide this angle into two equal parts:
- Each part will be \( \frac{\alpha}{2} \).

### Step 3: Identify the Triangles
We can consider two triangles:
1. Triangle \( OQK \) where \( O \) is the observer's eye, \( Q \) is the center of the balloon, and \( K \) is the point directly below \( Q \) on the horizontal line from \( O \).
2. Triangle \( OPQ \) where \( P \) is the point on the ground directly below \( O \).

### Step 4: Find \( OQ \) Using Sine
In triangle \( OQK \):
- The angle \( \angle OKQ = \frac{\alpha}{2} \).
- The radius \( OQ = r \).

Using the sine definition:
\[ \sin\left(\frac{\alpha}{2}\right) = \frac{KQ}{OQ} \]
Here, \( KQ \) is the radius \( r \), thus:
\[ KQ = OQ \cdot \sin\left(\frac{\alpha}{2}\right) \]
Since \( OQ = r \):
\[ KQ = r \cdot \sin\left(\frac{\alpha}{2}\right) \]

### Step 5: Find \( OP \) Using Sine
In triangle \( OPQ \):
- The angle \( \angle OPQ = \beta \).
- The hypotenuse \( OQ \) is already found as \( r \cdot \cos\left(\frac{\alpha}{2}\right) \).

Using the sine definition again:
\[ \sin(\beta) = \frac{OP}{OQ} \]
Rearranging gives:
\[ OP = OQ \cdot \sin(\beta) \]
Substituting \( OQ \):
\[ OP = \left(r \cdot \cos\left(\frac{\alpha}{2}\right)\right) \cdot \sin(\beta) \]

### Step 6: Final Expression
Thus, the height of the center of the balloon \( OP \) is given by:
\[ OP = r \cdot \cos\left(\frac{\alpha}{2}\right) \cdot \sin(\beta) \]

### Summary
The height of the center of the balloon is:
\[ \boxed{r \cdot \cos\left(\frac{\alpha}{2}\right) \cdot \sin(\beta)} \]