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

Options

A. `36. 3`
B. `37.3m`
C. `38.3m`
D. `39.3 m`

Answer

To solve the problem, we need to find the distance between the two poles given their heights and the angle made by the line joining their tops with the horizontal.

### Step-by-Step Solution:

1. **Identify the heights of the poles**:
- Let the height of the first pole (AB) be \( h_1 = 10 \, \text{m} \).
- Let the height of the second pole (CD) be \( h_2 = 20 \, \text{m} \).

2. **Calculate the vertical distance between the tops of the poles**:
- The vertical distance (CE) between the tops of the two poles is given by:
\[ CE = h_2 - h_1 = 20 \, \text{m} - 10 \, \text{m} = 10 \, \text{m} \]

3. **Set up the right triangle**:
- We have a right triangle ACE where:
- CE is the opposite side (10 m),
- AE is the base (the distance we need to find),
- The angle at A (angle between the horizontal and line joining the tops) is \( 15^\circ \).

4. **Use the tangent function**:
- From trigonometry, we know:
\[ \tan(15^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{CE}{AE} \]
- Substituting the known values:
\[ \tan(15^\circ) = \frac{10}{AE} \]

5. **Rearranging to find AE**:
- Rearranging the equation gives:
\[ AE = \frac{10}{\tan(15^\circ)} \]

6. **Calculate \( \tan(15^\circ) \)**:
- The value of \( \tan(15^\circ) \) can be calculated or looked up. It is approximately:
\[ \tan(15^\circ) \approx 0.2679 \]

7. **Substituting to find AE**:
- Now substituting the value of \( \tan(15^\circ) \):
\[ AE \approx \frac{10}{0.2679} \approx 37.3 \, \text{m} \]

8. **Final answer**:
- The distance between the two poles is approximately \( 37.3 \, \text{m} \).