If in `Delta ABC`, sides a, b, c are in A.P. then

Options

A. `B gt 60^(@)`
B. `B lt 60^(@)`
C. `B le 60^(@)`
D. `B = |A - C|`

Explanation

To solve the problem, we need to analyze the conditions given in triangle ABC where the sides a, b, and c are in Arithmetic Progression (A.P.).

### Step-by-Step Solution:

1. **Understanding the Condition**:
Since the sides a, b, and c are in A.P., we can express this as:
\[ 2b = a + c \]

2. **Using the Sine Rule**:
According to the sine rule, we can write:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
From this, we can express the sides in terms of the angles:
\[ a = 2R \sin A, \quad b = 2R \sin B, \quad c = 2R \sin C \]
where R is the circumradius of triangle ABC.

3. **Substituting in the A.P. Condition**:
Substitute the expressions for a and c into the A.P. condition:
\[ 2(2R \sin B) = 2R \sin A + 2R \sin C \]
Dividing through by \(2R\) (assuming \(R \neq 0\)):
\[ 2 \sin B = \sin A + \sin C \]

4. **Using the Sine Addition Formula**:
We can use the sine addition formula:
\[ \sin A + \sin C = 2 \sin \left(\frac{A + C}{2}\right) \cos \left(\frac{A - C}{2}\right) \]
Since \(A + B + C = 180^\circ\), we have:
\[ A + C = 180^\circ - B \]
Thus:
\[ \sin A + \sin C = 2 \sin \left(90^\circ - \frac{B}{2}\right) \cos \left(\frac{A - C}{2}\right) = 2 \cos \left(\frac{B}{2}\right) \cos \left(\frac{A - C}{2}\right) \]

5. **Setting Up the Equation**:
Now we can equate:
\[ 2 \sin B = 2 \cos \left(\frac{B}{2}\right) \cos \left(\frac{A - C}{2}\right) \]
Dividing both sides by 2:
\[ \sin B = \cos \left(\frac{B}{2}\right) \cos \left(\frac{A - C}{2}\right) \]

6. **Analyzing the Values**:
The maximum value of \(\cos\) is 1, thus:
\[ \sin B \leq \cos \left(\frac{A - C}{2}\right) \]
Since \(\sin B\) has a maximum value of 1, we can conclude:
\[ B \leq 60^\circ \]

### Conclusion:
Therefore, we conclude that if the sides a, b, and c of triangle ABC are in A.P., then:
\[ \angle B \leq 60^\circ \]