Question

With usual notations in \(△\)ABC, if \(bcos^2\frac{C}{2}+ccos^2\frac{B}{2} = \frac{3a}{2}\) then

• A. \(a,b,c\) are in A.P.
• B. \(a,c,b\) are in A.P.
• C. \(b,a,c\) are in A.P.
• D. \(a,b,c\) are in G.P

Hint:

Use cos^2(x/2) = (1 + cos x)/2 and the projection formula a = b cos C + c cos B.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Half-angle identity: \(\cos^2\dfrac{\theta}{2} = \dfrac{1 + \cos\theta}{2}\). Projection formula: \(a = b\cos C + c\cos B\).

Step 2: Rewrite the given condition:
\[ \frac{b(1 + \cos C)}{2} + \frac{c(1 + \cos B)}{2} = \frac{3a}{2} \]
\[ b + c + (b\cos C + c\cos B) = 3a \]

Step 3: Use the projection formula:
\(b\cos C + c\cos B = a\), so \(b + c + a = 3a\), giving \(b + c = 2a\).

Step 4: Interpret:
\(a = \dfrac{b + c}{2}\) means \(a\) is the arithmetic mean of \(b\) and \(c\), so \(b, a, c\) are in A.P. This is option (C). Option (A) would need \(b = \tfrac{a+c}{2}\), and (B) would need \(c = \tfrac{a+b}{2}\).

Final Answer:
b + c = 2a, so b, a, c are in A.P. \[ \boxed{\text{(C) }b,a,c\ \text{in A.P.}} \]