Step 1: Understanding the Concept:
We are given two sides of a triangle and the included angle, and asked to find the third side. This is a direct application of the Law of Cosines.
{Note: The side length \( \sqrt{3}-2 \) is mathematically negative (\( \approx -0.268 \)), which is physically impossible for a real triangle. However, treating this purely as an algebraic problem with the Cosine Rule yields a consistent result from the options.}
Step 2: Key Formula or Approach:
Law of Cosines: \( c^2 = a^2 + b^2 - 2ab \cos(C) \), where \( a \) and \( b \) are two sides, \( C \) is the included angle, and \( c \) is the third side.
Step 3: Detailed Explanation:
Let \( a = \sqrt{3} - 2 \), \( b = \sqrt{3} + 2 \), and angle \( C = 60^\circ \).
Using the Law of Cosines:
\[ c^2 = a^2 + b^2 - 2ab \cos(60^\circ) \]
We know \( \cos(60^\circ) = \frac{1}{2} \).
\[ c^2 = a^2 + b^2 - 2ab \left(\frac{1}{2}\right) \]
\[ c^2 = a^2 + b^2 - ab \]
Now, let's calculate the components:
\[ a^2 = (\sqrt{3} - 2)^2 = 3 + 4 - 4\sqrt{3} = 7 - 4\sqrt{3} \]
\[ b^2 = (\sqrt{3} + 2)^2 = 3 + 4 + 4\sqrt{3} = 7 + 4\sqrt{3} \]
\[ ab = (\sqrt{3} - 2)(\sqrt{3} + 2) = (\sqrt{3})^2 - 2^2 = 3 - 4 = -1 \]
Substitute these values back into the equation for \( c^2 \):
\[ c^2 = (7 - 4\sqrt{3}) + (7 + 4\sqrt{3}) - (-1) \]
\[ c^2 = 14 + 1 \]
\[ c^2 = 15 \]
Since side length must be positive, we take the positive square root:
\[ c = \sqrt{15} \]
Step 4: Final Answer:
The third side is \( \sqrt{15} \) units.