To solve this problem, let’s go through each step of the person’s movement carefully.

1. First Movement (from A to B):

   • The person starts at point $A$ and walks 15 meters east to reach point $B$.

2. Second Movement (120° Left Turn from B to C):

   • From point $B$, the person takes a 120° left turn and walks 15 meters. This turn means they are moving 120° counterclockwise from their eastward direction.
   • A 120° turn left from east puts the person at an angle of 120° from the positive x-axis, pointing in a direction that is roughly northwest.
   • After moving 15 meters in this direction, they reach point $C$.

3. Third Movement (120° Left Turn to Cross One of the Given Points):

   • At point $C$, the person again takes a 120° left turn. This turn would orient them towards point $A$, effectively making them return along the third side of an equilateral triangle.

Visualizing the Path as an Equilateral Triangle

The person’s movements trace an equilateral triangle with side lengths of 15 meters each. Since the person returns to point $A$ after the third movement, we know that points $A$, $B$, and $C$ form this equilateral triangle.

Calculating the Shortest Distance Between Point $C$ and Point $A$

In an equilateral triangle, the shortest distance between any vertex and the opposite side is the altitude of the triangle.

The formula for the altitude $h$ of an equilateral triangle with side length $s$ is: $h = \frac{\sqrt{3}}{2} s$ Substitute $s = 15$ meters: $h = \frac{\sqrt{3}}{2} \times 15 = \frac{15\sqrt{3}}{2} \approx 12.99 \, \text{meters}$

Answer

The shortest distance from point $C$ to point $A$ is approximately: $15\sqrt{3}/2 \approx 12.99 \text{ m}$

Since none of the provided options (15√2 m, 15 m, or 5√5 m) exactly match this answer, there may be a misalignment in options. However, the calculated shortest distance based on the triangle geometry is approximately 12.99 meters.

Would you like further clarification on this calculation?

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1. How is the altitude of an equilateral triangle derived?
2. Can we use trigonometry to verify this distance?
3. Why is an equilateral triangle formed in this scenario?
4. How would the path differ with a different angle of turn?
5. What if the sides had different lengths?

Tip: Visualizing problems involving directional turns on graph paper or with diagrams can simplify angle-related pathfinding.