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: Use the cosine rule form:
$\cos^2\tfrac C2 = \dfrac{s(s-c)}{ab}$ and $\cos^2\tfrac B2 = \dfrac{s(s-b)}{ac}$.

Step 2: Substitute:
$b\cdot\dfrac{s(s-c)}{ab} + c\cdot\dfrac{s(s-b)}{ac} = \dfrac{s}{a}\big[(s-c)+(s-b)\big] = \dfrac sa\cdot a = s$.

Step 3: Solve:
So $s = \tfrac{3a}{2}$, i.e. $a+b+c = 3a$ and $b + c = 2a$. Thus $b, a, c$ are in A.P., option (C).

Final Answer:
The sides b, a, c are in A.P. \[ \boxed{\text{(C) }b,a,c\ \text{in A.P.}} \]