A vertical tower stands on a horizontal plane and is surmounted by a flagstaff of height 5m. From a point on the ground the angles of elevation of the top and bottom of the flagstaff are `60^(@)` and `30^(@)` respectively. Find the height of the tower and the distance of the point from the tower. (Take `sqrt(3)=1.732`)

Answer

To solve the problem step by step, we will use trigonometry, specifically the tangent function, to find the height of the tower and the distance from the point to the tower.

### Step 1: Define the variables
Let:
- \( h \) = height of the tower (in meters)
- \( x \) = distance from the point on the ground to the base of the tower (in meters)

### Step 2: Analyze the first triangle (bottom of the flagstaff)
From the point on the ground, the angle of elevation to the bottom of the flagstaff (which is the top of the tower) is \( 30^\circ \).

Using the tangent function:
\[ \tan(30^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{x} \]
Given that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we can write:
\[ \frac{1}{\sqrt{3}} = \frac{h}{x} \]
Rearranging gives us:
\[ h = \frac{x}{\sqrt{3}} \quad \text{(1)} \]

### Step 3: Analyze the second triangle (top of the flagstaff)
The angle of elevation to the top of the flagstaff is \( 60^\circ \). The height to the top of the flagstaff is \( h + 5 \) (where 5 m is the height of the flagstaff).

Using the tangent function again:
\[ \tan(60^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h + 5}{x} \]
Given that \( \tan(60^\circ) = \sqrt{3} \), we can write:
\[ \sqrt{3} = \frac{h + 5}{x} \]
Rearranging gives us:
\[ h + 5 = \sqrt{3} \cdot x \quad \text{(2)} \]

### Step 4: Substitute equation (1) into equation (2)
From equation (1), we have:
\[ x = h \sqrt{3} \]
Substituting this into equation (2):
\[ h + 5 = \sqrt{3} \cdot (h \sqrt{3}) \]
This simplifies to:
\[ h + 5 = 3h \]
Rearranging gives:
\[ 5 = 3h - h \]
\[ 5 = 2h \]
Thus:
\[ h = \frac{5}{2} = 2.5 \text{ meters} \]

### Step 5: Find the distance \( x \)
Now, substituting \( h = 2.5 \) back into equation (1):
\[ x = h \sqrt{3} = 2.5 \cdot \sqrt{3} \]
Using \( \sqrt{3} \approx 1.732 \):
\[ x = 2.5 \cdot 1.732 \approx 4.33 \text{ meters} \]

### Final Results
- Height of the tower \( h = 2.5 \) meters
- Distance from the point to the tower \( x \approx 4.33 \) meters