From a point 20 m away from the foot of a tower, the angle of elevation of the top of the tower is `30^@`. Find the height of the tower.

Answer

To find the height of the tower, we can use the concept of trigonometry, specifically the tangent function, which relates the angle of elevation to the height of the tower and the distance from the tower.

### Step-by-Step Solution:

1. **Identify the elements of the problem**:
- Let \( AB \) be the height of the tower.
- Let \( C \) be the point from which the angle of elevation is measured.
- The distance from point \( C \) to the foot of the tower \( A \) is \( 20 \, \text{m} \).
- The angle of elevation \( \angle ACB \) is \( 30^\circ \).

2. **Set up the tangent function**:
- In triangle \( ABC \), we can use the tangent of the angle of elevation:
\[ \tan(\angle ACB) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{AC} \]
- Here, \( AB \) is the height of the tower, and \( AC \) is the distance from point \( C \) to the foot of the tower, which is \( 20 \, \text{m} \).

3. **Substitute the known values**:
- We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \).
- Therefore, we can write:
\[ \tan(30^\circ) = \frac{AB}{20} \]
- Substituting the value of \( \tan(30^\circ) \):
\[ \frac{1}{\sqrt{3}} = \frac{AB}{20} \]

4. **Solve for \( AB \)**:
- Rearranging the equation to find \( AB \):
\[ AB = 20 \cdot \frac{1}{\sqrt{3}} = \frac{20}{\sqrt{3}} \, \text{m} \]

5. **Rationalize the denominator (optional)**:
- To express the height in a more standard form, we can rationalize the denominator:
\[ AB = \frac{20 \cdot \sqrt{3}}{3} \, \text{m} \]

6. **Final Answer**:
- The height of the tower is:
\[ AB \approx 11.55 \, \text{m} \quad (\text{using } \sqrt{3} \approx 1.732) \]