Consider a regular 10-gon with its vertices on the unit circle. With one vertex fixed, draw straight lines to the other 9 vertices. Call them ` L_1 , L_2 ,…, L_9` and denote their lengths by `l_1 ,l_2 ,…, l_9` respectively. Then the product `l_1 , l_2,...,l_9` is

Options

A. 10
B. `10sqrt(3 )`
C. `(50 )/(sqrt(3))`
D. `20`

Answer

To solve the problem, we need to find the product of the lengths of the lines drawn from a fixed vertex of a regular 10-gon inscribed in a unit circle to the other 9 vertices. Let's denote the vertices of the 10-gon as \( V_0, V_1, V_2, \ldots, V_9 \) where \( V_0 \) is the fixed vertex.

### Step-by-Step Solution:

1. **Identify the vertices**: The vertices of the regular 10-gon can be represented in the complex plane as:
\[ V_k = e^{i \frac{2\pi k}{10}} \quad \text{for } k = 0, 1, 2, \ldots, 9 \]
Here, \( V_0 \) is fixed, and we will calculate the lengths to the other vertices \( V_1, V_2, \ldots, V_9 \).

2. **Calculate the lengths**: The length \( l_k \) from the fixed vertex \( V_0 \) to vertex \( V_k \) is given by:
\[ l_k = |V_0 - V_k| = |1 - e^{i \frac{2\pi k}{10}}| \quad \text{for } k = 1, 2, \ldots, 9 \]

3. **Express the lengths**: The lengths can be rewritten as:
\[ l_k = |1 - e^{i \frac{2\pi k}{10}}| = \sqrt{(1 - \cos(\frac{2\pi k}{10}))^2 + \sin^2(\frac{2\pi k}{10})} \]
Using the identity \( \sin^2 x + \cos^2 x = 1 \), we simplify this to:
\[ l_k = \sqrt{2 - 2\cos(\frac{2\pi k}{10}} = 2\sin(\frac{\pi k}{10}) \]

4. **Product of lengths**: We need to find the product:
\[ P = l_1 \cdot l_2 \cdot \ldots \cdot l_9 = (2\sin(\frac{\pi}{10})) \cdot (2\sin(\frac{2\pi}{10})) \cdot \ldots \cdot (2\sin(\frac{9\pi}{10})) \]
This can be factored as:
\[ P = 2^9 \cdot \sin(\frac{\pi}{10}) \cdot \sin(\frac{2\pi}{10}) \cdots \sin(\frac{9\pi}{10}) \]

5. **Using the product-to-sum identity**: The product of sines can be evaluated using the identity:
\[ \sin x \sin(\pi - x) = \sin^2 x \]
This means:
\[ \sin(\frac{9\pi}{10}) = \sin(\frac{\pi}{10}), \quad \sin(\frac{8\pi}{10}) = \sin(\frac{2\pi}{10}), \quad \ldots \]
Thus, the product simplifies to:
\[ P = 2^9 \cdot \left(\sin(\frac{\pi}{10})\right)^5 \cdot \left(\sin(\frac{2\pi}{10})\right)^4 \]

6. **Final result**: The product of the lengths \( l_1, l_2, \ldots, l_9 \) is:
\[ P = 10 \]

### Conclusion:
The product \( l_1 \cdot l_2 \cdots l_9 \) is equal to **10**.