Question

In \(\triangle \mathrm{ABC}\), with usual notations, if \(\cos \frac{B}{2}=\sqrt{\frac{c+a}{2 a}}\), then \(a^2=\)

• A. \(\mathrm{b}^{2}-\mathrm{c}^{2}\)
• B. \(\mathrm{b}+\mathrm{c}\)
• C. \(\mathrm{b}^{2}+\mathrm{c}^{2}\)
• D. \(\mathrm{b}-\mathrm{c}\)

Hint:

Use cos squared of B/2 in terms of the semi-perimeter s and the sides.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
In \(\triangle ABC\), \(\cos^2\dfrac B2=\dfrac{s(s-b)}{ac}\), where \(s=\dfrac{a+b+c}{2}\).

Step 2: Equate
Given \(\cos^2\dfrac B2=\dfrac{c+a}{2a}\), so \(\dfrac{s(s-b)}{ac}=\dfrac{c+a}{2a}\).
This gives \(s(s-b)=\dfrac{c(c+a)}{2}\).

Step 3: Simplify
\(s(s-b)=\dfrac{(a+c)^2-b^2}{4}\), so \((a+c)^2-b^2=2c(c+a)\).
\(a^2+2ac+c^2-b^2=2c^2+2ac\Rightarrow a^2=b^2+c^2\).

Final Answer:
\(a^2=b^2+c^2\). \[ \boxed{\text{(C) }\text{b}^{2}+\text{c}^{2}} \]