A stationary balloon is observed from three points A, B and C on the plane ground and is found point is `60^(@)`. If `angleABC=30^(@)` and AC = 5 meters, then the height of the balloon from the ground is

Options

A. `5sqrt2` meters
B. `(5)/(4)sqrt3` meters
C. `(5)/(sqrt3)` meters
D. `(5)/(4sqrt3)` meters

Answer

To solve the problem, we need to find the height of the balloon (denoted as \( h \)) from the ground, given the angles and distances involved.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- We have three points \( A \), \( B \), and \( C \) on the ground.
- The angle of elevation from each point to the balloon is \( 60^\circ \).
- The angle \( \angle ABC = 30^\circ \).
- The distance \( AC = 5 \) meters.

2. **Identify the Right Triangle**:
- The height of the balloon can be represented as the perpendicular from point \( P \) (the balloon) to the line \( AC \).
- Let \( PQ \) be the height of the balloon, which is \( h \).

3. **Using Trigonometric Ratios**:
- In triangle \( PAQ \) (where \( Q \) is the foot of the perpendicular from \( P \) to line \( AC \)), we can use the cotangent function:
\[ \cot(60^\circ) = \frac{h}{PA} \]
- We know that \( \cot(60^\circ) = \frac{1}{\sqrt{3}} \), thus:
\[ h = PA \cdot \cot(60^\circ) = PA \cdot \frac{1}{\sqrt{3}} \]

4. **Finding \( PA \)**:
- Since \( P \) is the circumcenter of triangle \( ABC \), the distances \( PA \), \( PB \), and \( PC \) are equal. Let this common distance be \( R \).
- Therefore, we have:
\[ PA = R = \frac{h}{\sqrt{3}} \]

5. **Using the Sine Rule in Triangle \( ABC \)**:
- In triangle \( ABC \):
\[ \sin(30^\circ) = \frac{AC}{BC} \]
- Given \( AC = 5 \):
\[ \sin(30^\circ) = \frac{1}{2} \Rightarrow BC = \frac{AC}{\sin(30^\circ)} = \frac{5}{\frac{1}{2}} = 10 \]

6. **Relating \( BC \) to the Circumradius**:
- The circumradius \( R \) is half of \( BC \) since \( P \) is the circumcenter:
\[ R = \frac{BC}{2} = \frac{10}{2} = 5 \]

7. **Substituting Back to Find \( h \)**:
- Now substituting \( R = 5 \) into the equation \( R = \frac{h}{\sqrt{3}} \):
\[ 5 = \frac{h}{\sqrt{3}} \Rightarrow h = 5\sqrt{3} \]

### Final Answer:
The height of the balloon from the ground is \( h = 5\sqrt{3} \) meters.