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

Options

A. `10 ( sqrt3 -1 ) m`
B. `5 (4 + 2 sqrt3) m `
C. `20 ( sqrt3 +1 )`
D. `10 ( sqrt3 + 1) m`

Answer

To solve the problem of finding the distance between two poles of heights 10 m and 20 m, where the line joining their tips makes an angle of 15 degrees with the horizontal, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have two poles: one is 10 m high (let's call it Pole A) and the other is 20 m high (Pole B).
- The difference in height between the two poles is \(20 m - 10 m = 10 m\).

2. **Draw the Diagram**:
- Draw two vertical lines representing the poles.
- Mark the height of Pole A (10 m) and Pole B (20 m).
- Draw a line connecting the tips of the two poles, which makes an angle of 15 degrees with the horizontal.

3. **Identify the Right Triangle**:
- The vertical difference between the two poles (10 m) acts as the opposite side of the right triangle.
- Let \(d\) be the horizontal distance between the two poles, which is the base of the triangle.

4. **Apply the Tangent Function**:
- According to the tangent function in trigonometry:
\[ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} \]
- Here, \(\theta = 15^\circ\), the opposite side is 10 m, and the adjacent side is \(d\).
- Therefore, we can write:
\[ \tan(15^\circ) = \frac{10}{d} \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ d = \frac{10}{\tan(15^\circ)} \]

6. **Calculate \(\tan(15^\circ)\)**:
- We can use the tangent subtraction formula:
\[ \tan(15^\circ) = \tan(45^\circ - 30^\circ) = \frac{\tan(45^\circ) - \tan(30^\circ)}{1 + \tan(45^\circ) \tan(30^\circ)} \]
- Knowing that \(\tan(45^\circ) = 1\) and \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\), we substitute:
\[ \tan(15^\circ) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \]

7. **Rationalize the Tangent**:
- Multiply the numerator and denominator by \(\sqrt{3} - 1\):
\[ \tan(15^\circ) = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{3 - 2\sqrt{3} + 1}{2} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \]

8. **Substitute Back to Find \(d\)**:
- Now substitute \(\tan(15^\circ)\) back into the equation for \(d\):
\[ d = \frac{10}{2 - \sqrt{3}} \]

9. **Rationalize \(d\)**:
- Multiply the numerator and denominator by \(2 + \sqrt{3}\):
\[ d = \frac{10(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})} = \frac{10(2 + \sqrt{3})}{4 - 3} = 10(2 + \sqrt{3}) \]

10. **Final Answer**:
- Thus, the distance \(d\) between the two poles is:
\[ d = 10(2 + \sqrt{3}) \text{ meters} \]