A pole of height h stands at one corner of a park of the shape of an equilateral triangle. If `alpha` is the angle which the pole subtends at the mid point of the opposite side, the length of each side of the park is

Options

A. `sqrt(3)/2 h cot alpha`
B. `2/sqrt(3)h cot alpha`
C. `sqrt(3)/2 h tan alpha`
D. `2/sqrt(3) h tan alpha`

Answer

To solve the problem, we need to find the length of each side of the equilateral triangle park, given the height of the pole (h) and the angle (α) that the pole subtends at the midpoint of the opposite side.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- Let the equilateral triangle be ABC, where A is the vertex with the pole of height h, and BC is the opposite side.
- Let M be the midpoint of side BC.
- The angle α is the angle subtended by the pole at point M.

2. **Setting Up the Triangle**:
- The height of the pole from point A to point M is h.
- The distance from point A to point M is the vertical height h, and the horizontal distance from M to the foot of the pole (point A) can be denoted as \( x \).

3. **Using Trigonometry**:
- In triangle AMB, we can use the tangent function:
\[ \tan(\alpha) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{x} \]
- Rearranging gives us:
\[ x = \frac{h}{\tan(\alpha)} \]

4. **Finding the Length of the Side of the Triangle**:
- The length of side BC (denoted as L) can be related to the distance x. Since M is the midpoint of BC, we can express L in terms of x:
\[ L = 2x \]
- Substituting for x from the previous step:
\[ L = 2 \left( \frac{h}{\tan(\alpha)} \right) = \frac{2h}{\tan(\alpha)} \]

5. **Final Expression**:
- We can also express \(\tan(\alpha)\) using the cotangent function:
\[ \tan(\alpha) = \frac{1}{\cot(\alpha)} \]
- Therefore, we can rewrite L as:
\[ L = 2h \cdot \cot(\alpha) \]

### Conclusion:
The length of each side of the park (equilateral triangle) is given by:
\[ L = \frac{2h}{\tan(\alpha)} \quad \text{or} \quad L = 2h \cdot \cot(\alpha) \]