On a ground , there is a vertical tower with a flagpole on its top . At a point 9 metre away from the foot of ther tower , the angles of elevation of the top and bottom of the flagpole are `60^(@)and30^(@)` respectively . The height of the flagpole is :

Options

A. a) `5sqrt(3)` metre
B. b) `6sqrt(3)` metre
C. c) `6sqrt(2)` metre
D. d) `6sqrt(5)` metre

Answer

To solve the problem, we need to find the height of the flagpole on top of the tower using the given angles of elevation from a point 9 meters away from the base of the tower.

### Step-by-Step Solution:

1. **Draw the Diagram**:
- Let the point on the ground where the observer is standing be point D.
- Let point B be the base of the tower, point C be the top of the tower, and point A be the top of the flagpole.
- The distance from point D to point B is 9 meters.
- The angle of elevation to point A (top of the flagpole) is 60 degrees, and the angle of elevation to point C (top of the tower) is 30 degrees.

2. **Identify the Heights**:
- Let the height of the tower (BC) be \( h_1 \).
- Let the height of the flagpole (CA) be \( h_2 \).
- The total height of the flagpole plus the tower will be \( h_1 + h_2 \).

3. **Using Trigonometry for the Tower**:
- For triangle BCD (where D is the observer's position):
\[ \tan(30^\circ) = \frac{h_1}{9} \]
- From the tangent value:
\[ h_1 = 9 \cdot \tan(30^\circ) = 9 \cdot \frac{1}{\sqrt{3}} = \frac{9}{\sqrt{3}} = 3\sqrt{3} \text{ meters} \]

4. **Using Trigonometry for the Flagpole**:
- For triangle ACD:
\[ \tan(60^\circ) = \frac{h_1 + h_2}{9} \]
- Substituting \( h_1 \):
\[ \tan(60^\circ) = \sqrt{3} \]
\[ \sqrt{3} = \frac{3\sqrt{3} + h_2}{9} \]
- Rearranging gives:
\[ 9\sqrt{3} = 3\sqrt{3} + h_2 \]
- Solving for \( h_2 \):
\[ h_2 = 9\sqrt{3} - 3\sqrt{3} = 6\sqrt{3} \text{ meters} \]

5. **Final Answer**:
- The height of the flagpole is \( 6\sqrt{3} \) meters.