Question

In \(\triangle ABC\) with usual notations, if \(a,c,b\) are in Arithmetic Progression (A.P.), then the value of \(\frac{a\cos B-b\cos A}{a-b}\) is:

• A. \(0\)
• B. \(1\)
• C. \(2\)
• D. \(3\)

Hint:

Use the cosine rule for cos A and cos B, then use 2c = a + b.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
By the cosine rule, \(a\cos B=\dfrac{a^2+c^2-b^2}{2c}\) and \(b\cos A=\dfrac{b^2+c^2-a^2}{2c}\).

Step 2: Subtract
\(a\cos B-b\cos A=\dfrac{2a^2-2b^2}{2c}=\dfrac{a^2-b^2}{c}\).

Step 3: Divide
\(\dfrac{a\cos B-b\cos A}{a-b}=\dfrac{a+b}{c}\).

Step 4: Use the AP Condition
Since \(a,c,b\) are in AP, \(2c=a+b\), so the value is \(\dfrac{2c}{c}=2\), option (C).

Final Answer:
The value is 2. \[ \boxed{\text{(C) }2} \]