Question

In a $\triangle ABC$, $a = 1$, $b = \sqrt{3}$ and $\angle C = \dfrac{\pi}{6}$. Then the measure of the third side $c =$}

• A. 4
• B. 3
• C. 1
• D. 2

Hint:

When two sides and the included angle are known (SAS), the Cosine Rule is the direct and only formula needed to find the third side.

Answer 1

The Correct Option is C

Step 1: Concept
Apply the Cosine Rule: $c^2 = a^2 + b^2 - 2ab\cos C$.

Step 2: Meaning
Here $a = 1$, $b = \sqrt{3}$, $C = \dfrac{\pi}{6} = 30^\circ$, so $\cos C = \dfrac{\sqrt{3}}{2}$.

Step 3: Analysis
Substituting: \[c^2 = 1^2 + (\sqrt{3})^2 - 2(1)(\sqrt{3})\cdot\frac{\sqrt{3}}{2} = 1 + 3 - 3 = 1.\]

Step 4: Conclusion
Therefore $c = 1$. Final Answer: (C)