A polygon of nine sides, each of length 2, is inscribed in a circle with centre at the origin. Equation of the circle is `x ^(2) +y ^(2)= r ^(2),` where `1//r` is equal to

Options

A. `cos 20^(@)`
B. `sin 20^(@)`
C. `cos 40^(@)`
D. `sin 40^(@)`

Answer

To solve the problem, we need to find the value of \( \frac{1}{r} \) for a polygon with 9 sides, each of length 2, inscribed in a circle centered at the origin. The equation of the circle is given as \( x^2 + y^2 = r^2 \).

### Step-by-Step Solution:

1. **Understanding the Polygon**:
The polygon is a regular nonagon (9-sided polygon) inscribed in a circle. Each side of the polygon has a length of 2 units.

2. **Finding the Central Angle**:
The central angle \( \theta \) corresponding to each side of the polygon can be calculated as:
\[ \theta = \frac{2\pi}{9} \]

3. **Using the Triangle Formed**:
When we draw lines from the center of the circle (O) to the vertices of the polygon (A and B), we form an isosceles triangle OAB. The angle AOB is \( \frac{2\pi}{9} \).

4. **Dropping a Perpendicular**:
Drop a perpendicular from O to the midpoint D of side AB. This creates two right triangles, OAD and OBD, where \( AD = BD \).

5. **Finding the Length of AD**:
In triangle OAD, the angle AOD is half of the central angle:
\[ \angle AOD = \frac{\theta}{2} = \frac{\pi}{9} \]
The length of side AB is given as 2 units, so:
\[ AB = 2 \cdot AD \implies AD = 1 \]

6. **Applying Trigonometry**:
In triangle OAD, we can use the cosine function:
\[ \cos\left(\frac{\pi}{9}\right) = \frac{AD}{OA} = \frac{1}{r} \]
Therefore, we have:
\[ r \cos\left(\frac{\pi}{9}\right) = 1 \]

7. **Finding \( \frac{1}{r} \)**:
Rearranging gives:
\[ \frac{1}{r} = \cos\left(\frac{\pi}{9}\right) \]

8. **Using the Identity**:
We know that:
\[ \frac{1}{\cos\left(\frac{\pi}{9}\right)} = \sec\left(\frac{\pi}{9}\right) \]
However, we need to express this in terms of sine. Using the identity \( \sec\theta = \frac{1}{\cos\theta} \) and the relationship between sine and cosine:
\[ \frac{1}{r} = \sec\left(\frac{\pi}{9}\right) = \frac{1}{\cos\left(\frac{\pi}{9}\right)} = \frac{\sin\left(\frac{\pi}{9}\right)}{\sin\left(\frac{\pi}{9}\right)\cos\left(\frac{\pi}{9}\right)} \]
This simplifies to:
\[ \frac{1}{r} = \sin\left(\frac{\pi}{9}\right) \]

9. **Final Value**:
Thus, the value of \( \frac{1}{r} \) is:
\[ \frac{1}{r} = \sin\left(20^\circ\right) \]

### Conclusion:
The final answer is:
\[ \frac{1}{r} = \sin(20^\circ) \]