Question

If in triangle ABC , with usual notations \(\sin \frac{\mathrm{A}}{2} \cdot \sin \frac{\mathrm{C}}{2}=\sin \frac{\mathrm{B}}{2}\) and 2 s is the perimeter of the triangle, then the value of \(s\) is

• A. 2 b
• B. b
• C. \(4b\)
• D. \(\frac{\mathrm{b}}{2}\)

Hint:

Use the half-angle formulas in terms of the semi-perimeter s and compare.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
\(\sin\frac A2=\sqrt{\frac{(s-b)(s-c)}{bc}}\), \(\sin\frac B2=\sqrt{\frac{(s-a)(s-c)}{ac}}\), \(\sin\frac C2=\sqrt{\frac{(s-a)(s-b)}{ab}}\).

Step 2: Product
\(\sin\frac A2\sin\frac C2=\frac{(s-b)\sqrt{(s-a)(s-c)}}{b\sqrt{ac}}\).

Step 3: Equate with sin(B/2)
\(\frac{(s-b)\sqrt{(s-a)(s-c)}}{b\sqrt{ac}}=\frac{\sqrt{(s-a)(s-c)}}{\sqrt{ac}}\).

Step 4: Solve
Cancelling gives \(\frac{s-b}b=1\Rightarrow s=2b\), option (A).

Final Answer:
The value of s is 2b. \[ \boxed{\text{(A) }2b} \]