A stationary balloon is observed from three points A, B and C on the plane groundand is found that its angle of elevation from each 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 step by step, we will use trigonometric relationships and properties of triangles.

### Step 1: Understand the Geometry
We have three points A, B, and C on the ground, and a balloon directly above point P. The angles of elevation from points A, B, and C to the balloon are all 60 degrees. The angle ABC is given as 30 degrees, and the distance AC is 5 meters.

### Step 2: Set Up the Triangle
Let:
- \( H \) be the height of the balloon from the ground.
- \( AC = 5 \) meters.
- \( \angle ABC = 30^\circ \).
- The angles of elevation from A and C to the balloon (point P) are \( 60^\circ \).

### Step 3: Use the Sine Rule in Triangle ABC
In triangle ABC, we can use the sine rule. The sine of angle ABC can be expressed as:
\[ \sin(30^\circ) = \frac{AC}{BC} \]
Where \( BC \) is the length from point B to point C. Since \( \sin(30^\circ) = \frac{1}{2} \), we have:
\[ \frac{1}{2} = \frac{5}{BC} \implies BC = 10 \text{ meters} \]

### Step 4: Identify the Relationship Between Points
Since the angles of elevation from A and C to the balloon are both \( 60^\circ \), we can denote the distances from A to P and C to P as \( PA \) and \( PC \) respectively. By the properties of triangles:
- \( PA = PC \) because the angles of elevation are the same.

### Step 5: Use the Tangent Function
For triangle APQ (where Q is the point directly below the balloon on the ground):
\[ \tan(60^\circ) = \frac{H}{PA} \]
Since \( \tan(60^\circ) = \sqrt{3} \), we can write:
\[ \sqrt{3} = \frac{H}{PA} \implies H = PA \cdot \sqrt{3} \]

### Step 6: Relate PA to the Radius
From the previous steps, we know that:
- The circumradius \( R \) of triangle ABC can be expressed as:
\[ R = \frac{AC}{2 \sin(30^\circ)} = \frac{5}{2 \cdot \frac{1}{2}} = 5 \text{ meters} \]
Since \( PA \) is equal to the circumradius \( R \):
\[ PA = 5 \text{ meters} \]

### Step 7: Substitute Back to Find H
Now substitute \( PA \) back into the equation for height \( H \):
\[ H = PA \cdot \sqrt{3} = 5 \cdot \sqrt{3} \]

### Final Answer
Thus, the height of the balloon from the ground is:
\[ H = 5\sqrt{3} \text{ meters} \]