Two poles are 10 m and 20 m high. The line joining their tops makes an angle of `15^(@)` with the horizontal. What is the approximate distance between the poles?

Options

A. `10(sqrt(3)-1)m`
B. `5(4+2sqrt(3))m`
C. `20(sqrt(3)+1)m`
D. `10(sqrt(3)+1)m`

Explanation

To solve the problem, we need to find the distance between two poles of heights 10 m and 20 m, where the line joining their tops makes an angle of 15 degrees with the horizontal.

### Step-by-Step Solution:

1. **Identify the heights of the poles**:
- Let the height of the first pole (A) be \( h_1 = 10 \) m.
- Let the height of the second pole (B) be \( h_2 = 20 \) m.

2. **Determine the height difference**:
- The height difference \( H \) between the two poles is given by:
\[ H = h_2 - h_1 = 20 \, \text{m} - 10 \, \text{m} = 10 \, \text{m} \]

3. **Set up the right triangle**:
- Let the distance between the bases of the two poles be \( x \).
- The line joining the tops of the poles forms the hypotenuse of a right triangle, where:
- One leg is the height difference \( H = 10 \) m.
- The other leg is the horizontal distance \( x \).

4. **Use the tangent function**:
- The angle \( \theta = 15^\circ \) is given, and we can use the tangent function:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{H}{x} \]
- Therefore, we have:
\[ \tan(15^\circ) = \frac{10}{x} \]

5. **Calculate \( x \)**:
- Rearranging the equation gives:
\[ x = \frac{10}{\tan(15^\circ)} \]
- Using the approximate value of \( \tan(15^\circ) \approx 0.2679 \):
\[ x \approx \frac{10}{0.2679} \approx 37.34 \, \text{m} \]

6. **Final Answer**:
- The approximate distance between the two poles is:
\[ \boxed{37.34 \, \text{m}} \]