Question

If two sides of a triangle are \( \sqrt{3} - 1 \) and \( \sqrt{3} + 1 \) units and their included angle is \( 60^\circ \), then the third side of the triangle is

• A. 15 units
• B. \( \sqrt{15} - 2 \) units
• C. \( \sqrt{15} + 2 \) units
• D. \( \sqrt{6} \) units

Hint:

Cosine rule is perfect for finding the third side when two sides and the included angle are known (SAS).

Answer 1

The Correct Option is D

Step 1: Concept
Use the Cosine Rule: $a^2 = b^2 + c^2 - 2bc \cos A$.

Step 2: Meaning
Let $b = \sqrt{3} - 1$, $c = \sqrt{3} + 1$, and $A = 60^\circ$.

Step 3: Analysis
$a^2 = (\sqrt{3}-1)^2 + (\sqrt{3}+1)^2 - 2(\sqrt{3}-1)(\sqrt{3}+1)\cos 60^\circ$. $a^2 = (3+1-2\sqrt{3}) + (3+1+2\sqrt{3}) - 2(3-1)(1/2)$. $a^2 = 8 - 2 = 6$.

Step 4: Conclusion
$a = \sqrt{6}$ units. Final Answer: (D)