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Explanation:

Since Δ ABC is an equilateral triangle with length of the side 2 km, so its altitude will be $\sqrt{3}$ km. 

The horizontal distance of D from A = half of AB + CD = 1 + 2 = 3 kms

The vertical distance of D from A = height of the triangle ABC = √3 kms

∴ D is 3 km east and $\sqrt{3}$ km north of A.

Hence, option (b).