A spherical balloon of radius 50 cm, subtends an angle of `60^(@)` at a man's eye when the elevation of its centre if `45^(@)`. Then the height of the centre of the ballon is

Options

A. `50sqrt(3)`cm
B. 50 m
C. `40sqrt(2)`cm
D. `50sqrt(2)` cm

Answer

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

### Step 1: Understand the Problem
We have a spherical balloon with a radius of 50 cm, which subtends an angle of 60° at a man's eye when the elevation of its center is 45°. We need to find the height of the center of the balloon from the man's eye level.

### Step 2: Define Variables
Let:
- \( r = 50 \) cm (radius of the balloon)
- \( D = 2r = 100 \) cm (diameter of the balloon)
- \( \alpha = 60^\circ \) (angle subtended at the man's eye)
- \( \beta = 45^\circ \) (angle of elevation to the center of the balloon)

### Step 3: Use the Properties of Tangents
From the point of view of the man, the tangents drawn to the balloon from his eye are equal in length. Let \( E \) be the man's eye level, \( O \) be the center of the balloon, and \( A \) and \( B \) be the points on the surface of the balloon where the tangents touch.

### Step 4: Set Up the Triangles
In triangles \( AOE \) and \( BOE \):
- \( AE = BE \) (tangents from point E)
- \( AO = BO = \frac{D}{2} = 50 \) cm (radius of the balloon)

### Step 5: Use Sine Rule in Triangle AOE
Using the sine rule in triangle \( AOE \):
\[ \sin\left(\frac{\alpha}{2}\right) = \frac{\frac{D}{2}}{OE} \]
Thus,
\[ OE = \frac{D/2}{\sin\left(\frac{\alpha}{2}\right)} = \frac{50}{\sin(30^\circ)} = \frac{50}{\frac{1}{2}} = 100 \text{ cm} \]

### Step 6: Use Sine Rule in Triangle OLE
Now, in triangle \( OLE \):
\[ \sin(\beta) = \frac{OL}{OE} \]
Thus,
\[ OL = OE \cdot \sin(\beta) \]
Substituting the value of \( OE \):
\[ OL = 100 \cdot \sin(45^\circ) = 100 \cdot \frac{1}{\sqrt{2}} = 50\sqrt{2} \text{ cm} \]

### Step 7: Find the Height of the Center of the Balloon
The height \( h \) of the center of the balloon from the man's eye level is:
\[ h = OL + r = 50\sqrt{2} + 50 \]

### Step 8: Final Calculation
Calculating \( 50\sqrt{2} \):
\[ h \approx 50 \cdot 1.414 + 50 \approx 70.7 + 50 = 120.7 \text{ cm} \]

### Conclusion
Thus, the height of the center of the balloon from the man's eye level is approximately \( 120.7 \) cm.