Two parallel chords of a circle of radius 2 are at a distance `sqrt(3)+1` apart. If the chords subtend at the centre, angle of `(pi)/(k)` and `(2pi)/(k)` where `k gt 0`. Then the value of [k] is

Answer

To solve the problem, we need to analyze the situation involving two parallel chords in a circle of radius 2, which are at a distance of \(\sqrt{3} + 1\) apart. The chords subtend angles of \(\frac{\pi}{k}\) and \(\frac{2\pi}{k}\) at the center of the circle.

### Step-by-step Solution:

1. **Understanding the Geometry**:
- Let the center of the circle be \(O\).
- Let the two parallel chords be \(AB\) and \(CD\).
- The distance between the chords \(AB\) and \(CD\) is given as \(\sqrt{3} + 1\).

2. **Setting Up the Angles**:
- The angle subtended by chord \(AB\) at the center \(O\) is \(\frac{\pi}{k}\).
- The angle subtended by chord \(CD\) at the center \(O\) is \(\frac{2\pi}{k}\).

3. **Using the Cosine Rule**:
- For chord \(AB\):
\[ OA = OB = r = 2 \]
The length of chord \(AB\) can be found using:
\[ AB = 2 \times OA \times \sin\left(\frac{\frac{\pi}{k}}{2}\right) = 2 \times 2 \times \sin\left(\frac{\pi}{2k}\right) = 4 \sin\left(\frac{\pi}{2k}\right) \]

- For chord \(CD\):
\[ CD = 2 \times OC \times \sin\left(\frac{\frac{2\pi}{k}}{2}\right) = 2 \times 2 \times \sin\left(\frac{\pi}{k}\right) = 4 \sin\left(\frac{\pi}{k}\right) \]

4. **Finding the Distance Between the Chords**:
- The distance between the two chords can be expressed as:
\[ d = 2 \cos\left(\frac{\pi}{2k}\right) + 2 \cos\left(\frac{\pi}{k}\right) \]
- Given that this distance equals \(\sqrt{3} + 1\), we can write:
\[ 2 \cos\left(\frac{\pi}{2k}\right) + 2 \cos\left(\frac{\pi}{k}\right) = \sqrt{3} + 1 \]
- Dividing the entire equation by 2 gives:
\[ \cos\left(\frac{\pi}{2k}\right) + \cos\left(\frac{\pi}{k}\right) = \frac{\sqrt{3} + 1}{2} \]

5. **Substituting for \(t = \cos\left(\frac{\pi}{2k}\right)\)**:
- Let \(t = \cos\left(\frac{\pi}{2k}\right)\), then:
\[ \cos\left(\frac{\pi}{k}\right) = 2t^2 - 1 \]
- Substituting this into the equation gives:
\[ t + (2t^2 - 1) = \frac{\sqrt{3} + 1}{2} \]
- Rearranging leads to:
\[ 2t^2 + t - 1 - \frac{\sqrt{3} + 1}{2} = 0 \]

6. **Solving the Quadratic Equation**:
- Multiply through by 2 to eliminate the fraction:
\[ 4t^2 + 2t - 2 - (\sqrt{3} + 1) = 0 \]
\[ 4t^2 + 2t - (\sqrt{3} + 3) = 0 \]
- Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ t = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 4 \cdot (-(\sqrt{3} + 3))}}{2 \cdot 4} \]
\[ t = \frac{-2 \pm \sqrt{4 + 16(\sqrt{3} + 3)}}{8} \]

7. **Finding \(k\)**:
- After solving for \(t\), we will find \(\frac{\pi}{2k} = \cos^{-1}(t)\) and subsequently find \(k\).

8. **Calculating the Greatest Integer Function**:
- Once we find \(k\), we will compute \([k]\), the greatest integer less than or equal to \(k\).