A man moves in an open field such that after moving 10 m on a straight line , he makes a sharp turn of `60^(@)` to his left. The total displacement just at the start of `8^(th)` turn is equal to

Options

A. 12 m
B. 15 m
C. 17.32 m
D. 14. 14 m

Answer

To solve the problem, we need to calculate the total displacement of the man after he has made 8 turns, each involving a movement of 10 meters followed by a 60-degree turn to the left.

### Step-by-Step Solution:

1. **Understanding the Movement**:
The man moves 10 meters in a straight line and then makes a 60-degree turn to his left. This process is repeated for a total of 8 segments.

2. **Identifying the Geometry**:
Each segment forms a triangle with the previous segment. After each turn, the man effectively creates an angle of 60 degrees between his current direction and the previous segment.

3. **Calculating the Displacement after Each Turn**:
To find the total displacement after the 8th turn, we can visualize the path as a series of vectors. The displacement after each segment can be calculated using the cosine rule.

4. **Using the Cosine Rule**:
The displacement vector after the first two segments (OA and AB) can be calculated as follows:
- Let \( OA = 10 \, m \) and \( AB = 10 \, m \).
- The angle between them is \( 60^\circ \).
- Using the cosine rule:
\[ OB^2 = OA^2 + AB^2 - 2 \cdot OA \cdot AB \cdot \cos(60^\circ) \]
- Substituting the values:
\[ OB^2 = 10^2 + 10^2 - 2 \cdot 10 \cdot 10 \cdot \frac{1}{2} \]
\[ OB^2 = 100 + 100 - 100 = 100 \]
- Therefore, \( OB = \sqrt{100} = 10 \, m \).

5. **Repeating the Process**:
This process continues for each segment, and we can observe that after every two segments, the displacement remains consistent due to the symmetry of the turns.

6. **Calculating Total Displacement after 8 Turns**:
After 8 turns, the man will have completed 8 segments of 10 meters each. Since the angle remains consistent, we can calculate the resultant displacement after 8 segments:
- The total displacement after 8 turns can be calculated as:
\[ R = n \cdot d \cdot \cos\left(\frac{n-1}{2} \cdot 60^\circ\right) \]
- Where \( n = 8 \) and \( d = 10 \, m \):
\[ R = 8 \cdot 10 \cdot \cos(210^\circ) \]
- Since \( \cos(210^\circ) = -\frac{\sqrt{3}}{2} \):
\[ R = 80 \cdot -\frac{\sqrt{3}}{2} = -40\sqrt{3} \approx -69.28 \, m \]
- The magnitude of the displacement is \( 40\sqrt{3} \, m \).

### Final Result:
The total displacement just at the start of the 8th turn is approximately \( 40\sqrt{3} \, m \) or about \( 69.28 \, m \).