A man stands on the ground at a point A, which is on the same horizontal plane as B, the foot of a vertical pole BC. The height of the pole is 10 m . The man's eye is 2 m above the ground . He observes the angle of elevation at C, The top of the pole as `x^(@)` where tan `x^(@) = (2)/(5)` . the distance AB (in metres) is

Options

A. 15 m
B. 18 m
C. 20 m
D. 16 m

Answer

To solve the problem step by step, we will use the information provided in the question and apply some basic trigonometry concepts.

### Step-by-Step Solution:

1. **Understand the Setup**:
- Let point A be where the man is standing.
- Point B is the foot of the vertical pole.
- Point C is the top of the pole.
- The height of the pole (BC) is 10 m, and the man’s eye level is 2 m above the ground.

2. **Determine the Effective Height**:
- The height of the pole above the man's eye level (CD) can be calculated as:
\[ CD = BC - AE = 10 \, \text{m} - 2 \, \text{m} = 8 \, \text{m} \]

3. **Use the Angle of Elevation**:
- The angle of elevation from the man's eyes to the top of the pole is given as \( x \) degrees, where \( \tan(x) = \frac{2}{5} \).

4. **Set Up the Tangent Ratio**:
- In triangle CED (where D is the point directly below C on the horizontal line from A), we can use the tangent function:
\[ \tan(x) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{CD}{AD} \]
- Here, \( CD = 8 \, \text{m} \) and \( AD \) is the distance we want to find (AB).

5. **Substitute the Values**:
- From the tangent ratio:
\[ \tan(x) = \frac{8}{AD} \]
- We know \( \tan(x) = \frac{2}{5} \), so we can write:
\[ \frac{2}{5} = \frac{8}{AD} \]

6. **Cross Multiply to Solve for AD**:
- Cross multiplying gives:
\[ 2 \cdot AD = 8 \cdot 5 \]
\[ 2 \cdot AD = 40 \]
\[ AD = \frac{40}{2} = 20 \, \text{m} \]

7. **Conclusion**:
- The distance \( AB \) is 20 meters.

### Final Answer:
The distance \( AB \) is **20 meters**.