Question

In \(\triangle ABC\), if \(b=\sqrt{3}+1\), \(c=\sqrt{3}-1\) and \(A=\frac{\pi }{3}\), then \(B-C=\ldots\)

• A. \(\frac{\pi }{18}\)
• B. \(\frac{\pi }{12}\)
• C. \(\frac{\pi }{4}\)
• D. \(\frac{\pi }{2}\)

Hint:

Use tan((B-C)/2) = ((b-c)/(b+c)) cot(A/2).

Answer 1

The Correct Option is D

Step 1: Write Napier's analogy
\(\tan\dfrac{B-C}{2}=\dfrac{b-c}{b+c}\cot\dfrac{A}{2}\)

Step 2: Find the ratio
\(b-c=2\) and \(b+c=2\sqrt3\), so \(\dfrac{b-c}{b+c}=\dfrac{1}{\sqrt3}\).

Step 3: Cotangent
\(\cot\dfrac{A}{2}=\cot\dfrac{\pi}{6}=\sqrt3\)

Step 4: Solve
\(\tan\dfrac{B-C}{2}=\dfrac1{\sqrt3}\cdot\sqrt3=1\), so \(\dfrac{B-C}{2}=\dfrac{\pi}{4}\) and \(B-C=\dfrac{\pi}{2}\).

Final Answer:
B minus C equals pi by 2. \[ \boxed{\text{(D) }\dfrac{\pi}{2}} \]