Question

In a triangle $ ABC $, if angles $ A $, $ B $, and $ C $ are in A.P., then $ \frac{a + c}{b} $ is equal to:

• A. \( \frac{2 \sin \frac{A - C}{2}}{2} \)
• B. \( 2 \cos \frac{A - C}{2} \)
• C. \( \frac{\cos \frac{A - C}{2}}{2} \)
• D. \( \sin \frac{A - C}{2} \)

Hint:

In triangles where angles are in A.P., use the Law of Sines and angle sum identities to simplify expressions. This often leads to a neat relation between the sides.

Answer 1

The Correct Option is B

We are given a triangle \( ABC \) where the angles \( A \), \( B \), and \( C \) are in A.P. This implies:
\[ A - B = B - C \quad \Rightarrow \quad 2B = A + C. \] Therefore, \[ B = \frac{A + C}{2}. \] Next, we need to find the expression for \( \frac{a + c}{b} \), where \( a \), \( b \), and \( c \) are the sides opposite to angles \( A \), \( B \), and \( C \), respectively. 
Step 1: Use the Law of Sines.
By the Law of Sines, we know: \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}. \] From this, we can express the sides in terms of the sine of the angles: \[ a = k \sin A, \quad b = k \sin B, \quad c = k \sin C, \] where \( k \) is the common constant, \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \). 
Step 2: Expression for \( \frac{a + c}{b} \).
We need to find: \[ \frac{a + c}{b} = \frac{k (\sin A + \sin C)}{k \sin B}. \] Simplifying: \[ \frac{a + c}{b} = \frac{\sin A + \sin C}{\sin B}. \] Using the sum of sines identity: \[ \sin A + \sin C = 2 \sin \left( \frac{A + C}{2} \right) \cos \left( \frac{A - C}{2} \right). \] Therefore, the expression becomes: \[ \frac{a + c}{b} = \frac{2 \sin \left( \frac{A + C}{2} \right) \cos \left( \frac{A - C}{2} \right)}{\sin B}. \] Since \( B = \frac{A + C}{2} \), we have: \[ \frac{a + c}{b} = 2 \cos \left( \frac{A - C}{2} \right). \] Thus, the correct answer is \( B \).