Abhinav and Brijesh started walking simultaneously @ 3 km/h and 4 km/h from the same point 'C' on two different paths which diverge from each other at an angle of `120^@`. They walked for 6 hours and they stopped at A and B respectively. What is the least possible distance between Abhinav and Brijesh when they are at A and B respectively.

Options

A. `36sqrt(37)` km
B. `6sqrt(37)` km
C. `12sqrt(37)` km
D. none of these

Answer

To find the least possible distance between Abhinav and Brijesh when they are at points A and B respectively, we can use the Law of Cosines in the triangle formed by their paths.

### Step-by-Step Solution:

1. **Determine the distances traveled by Abhinav and Brijesh:**
- Abhinav's speed = 3 km/h
- Brijesh's speed = 4 km/h
- Time = 6 hours
- Distance traveled by Abhinav (AC) = Speed × Time = 3 km/h × 6 h = 18 km
- Distance traveled by Brijesh (BC) = Speed × Time = 4 km/h × 6 h = 24 km

2. **Identify the angle between their paths:**
- The angle between the paths (∠ACB) = 120°

3. **Apply the Law of Cosines:**
The Law of Cosines states that for any triangle with sides a, b, and c opposite to angles A, B, and C respectively:
\[ c^2 = a^2 + b^2 - 2ab \cdot \cos(C) \]
Here, let:
- AC = 18 km (side a)
- BC = 24 km (side b)
- ∠ACB = 120° (angle C)
- AB = x (the distance we want to find)

Plugging in the values, we have:
\[ x^2 = 18^2 + 24^2 - 2 \cdot 18 \cdot 24 \cdot \cos(120°) \]

4. **Calculate the cosine of 120°:**
- \(\cos(120°) = -\frac{1}{2}\)

5. **Substituting the values into the equation:**
\[ x^2 = 18^2 + 24^2 - 2 \cdot 18 \cdot 24 \cdot \left(-\frac{1}{2}\right) \]
\[ x^2 = 324 + 576 + 18 \cdot 24 \]
\[ x^2 = 324 + 576 + 432 \]
\[ x^2 = 1332 \]

6. **Taking the square root to find x:**
\[ x = \sqrt{1332} \]
\[ x = \sqrt{4 \cdot 333} = 2\sqrt{333} \]

7. **Final Calculation:**
To simplify further, we can approximate \(\sqrt{333}\):
\[ \sqrt{333} \approx 18.25 \quad \text{(approximately)} \]
Thus,
\[ x \approx 2 \cdot 18.25 \approx 36.5 \text{ km} \]

### Conclusion:
The least possible distance between Abhinav and Brijesh when they are at points A and B respectively is approximately \(2\sqrt{333}\) km or about 36.5 km.