To determine how far Ramesh was from the starting point, we need to track his movements using coordinate geometry. We will break down his journey into segments, calculating his position after each turn and movement.

Ramesh's Journey: Tracing the Path

Ramesh's journey involves a series of movements in specific directions and then turns at certain angles. We can imagine his starting point as the origin (0, 0) on a 2D coordinate plane.

• First Movement: 15 m to the East.
• Second Movement: Turns left (North direction) and moves \(5\sqrt{3}\) m.
• Third Movement: Turns 120° right and moves \(10\sqrt{3}\) m.
• Fourth Movement: Turns 120° right again and moves \(20\sqrt{3}\) m.

Path Direction: Understanding Each Turn

Understanding the direction after each turn is crucial for accurate coordinate calculation. We will use standard angles where East is 0°, North is 90°, West is 180°, and South is 270° (or -90°).

1. Initial Direction: East (0°).
2. After First Turn (left): From East, turning left means turning counter-clockwise by 90°. So, the direction becomes North (90°).
3. After Second Turn (120° right): From North (90°), turning right means turning clockwise by 120°. The new direction angle is \(90^\circ - 120^\circ = -30^\circ\). A direction of -30° is equivalent to 330° (South-East quadrant).
4. After Third Turn (120° right): From the current direction (-30° or 330°), turning right means turning clockwise by 120°. The new direction angle is \(-30^\circ - 120^\circ = -150^\circ\). A direction of -150° is equivalent to 210° (South-West quadrant).

Coordinate Calculation: Mapping Ramesh's Movement

Let the starting point be \(O = (0, 0)\). We will calculate the coordinates after each segment of Ramesh's movement.

Movement 1: Eastward Displacement

• Description: Ramesh moves 15 m to the East.
• New Position (Point A): From \((0, 0)\), moving 15 m East means adding 15 to the x-coordinate.
  • \(x_A = 0 + 15 = 15\) m
  • \(y_A = 0\) m
• Coordinates of A: \((15, 0)\)

Movement 2: Northward Displacement

• Description: From point A, Ramesh turns left (North) and moves \(5\sqrt{3}\) m.
• New Position (Point B): From \((15, 0)\), moving \(5\sqrt{3}\) m North means adding \(5\sqrt{3}\) to the y-coordinate.
  • \(x_B = 15\) m
  • \(y_B = 0 + 5\sqrt{3} = 5\sqrt{3}\) m
• Coordinates of B: \((15, 5\sqrt{3})\)

Movement 3: First 120° Turn Analysis

• Description: From point B, Ramesh turns 120° right and moves \(10\sqrt{3}\) m. The current direction was North (90°). Turning 120° right means the new direction is \(90^\circ - 120^\circ = -30^\circ\).
• Displacement Components:
  • Change in x: \(d \cdot \cos(\theta) = 10\sqrt{3} \cdot \cos(-30^\circ) = 10\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 5 \cdot 3 = 15\) m
  • Change in y: \(d \cdot \sin(\theta) = 10\sqrt{3} \cdot \sin(-30^\circ) = 10\sqrt{3} \cdot (-\frac{1}{2}) = -5\sqrt{3}\) m
• New Position (Point C): Adding these changes to coordinates of B \((15, 5\sqrt{3})\).
  • \(x_C = 15 + 15 = 30\) m
  • \(y_C = 5\sqrt{3} + (-5\sqrt{3}) = 0\) m
• Coordinates of C: \((30, 0)\)

Movement 4: Second 120° Turn Analysis

• Description: From point C, Ramesh turns 120° right again and moves \(20\sqrt{3}\) m. The current direction was -30°. Turning 120° right means the new direction is \(-30^\circ - 120^\circ = -150^\circ\).
• Displacement Components:
  • Change in x: \(d \cdot \cos(\theta) = 20\sqrt{3} \cdot \cos(-150^\circ) = 20\sqrt{3} \cdot (-\frac{\sqrt{3}}{2}) = -10 \cdot 3 = -30\) m
  • Change in y: \(d \cdot \sin(\theta) = 20\sqrt{3} \cdot \sin(-150^\circ) = 20\sqrt{3} \cdot (-\frac{1}{2}) = -10\sqrt{3}\) m
• New Position (Point D): Adding these changes to coordinates of C \((30, 0)\).
  • \(x_D = 30 + (-30) = 0\) m
  • \(y_D = 0 + (-10\sqrt{3}) = -10\sqrt{3}\) m
• Coordinates of D: \((0, -10\sqrt{3})\)

Distance Calculation: Finding Final Position

Ramesh started at point \(O = (0, 0)\) and ended at point \(D = (0, -10\sqrt{3})\). The distance from the starting point is the straight-line distance between O and D.

Using the distance formula: \(D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

Here, \((x_1, y_1) = (0, 0)\) and \((x_2, y_2) = (0, -10\sqrt{3})\).

Distance \(OD = \sqrt{(0 - 0)^2 + (-10\sqrt{3} - 0)^2}\)

\(OD = \sqrt{(0)^2 + (-10\sqrt{3})^2}\)

\(OD = \sqrt{0 + (10^2 \cdot (\sqrt{3})^2)}\)

\(OD = \sqrt{100 \cdot 3}\)

\(OD = \sqrt{300}\)

To simplify \(\sqrt{300}\):

\(OD = \sqrt{100 \cdot 3} = \sqrt{100} \cdot \sqrt{3} = 10\sqrt{3}\) m

Therefore, Ramesh was \(10\sqrt{3}\) m from the starting point.

The final coordinates of Ramesh are \((0, -10\sqrt{3})\). The distance from the starting point \((0,0)\) is simply the magnitude of the y-coordinate, which is \(|-10\sqrt{3}| = 10\sqrt{3}\) m.