If the angle A,B,C of a `Delta ABC ` are in A.P then :-

Options

A. `c^2=a^2+b^2-ab`
B. `b^2=a^2+c^2-ac`
C. `c^2=a^2+b^2`
D. none of these

Answer

To solve the problem, we need to establish the relationship between the angles of triangle ABC when they are in Arithmetic Progression (AP). Let's denote the angles as follows:

- Let \( A = a \)
- Let \( B = b \)
- Let \( C = c \)

### Step 1: Establish the condition for angles in AP
If the angles \( A, B, C \) are in AP, then the middle angle \( B \) can be expressed as:
\[ B = \frac{A + C}{2} \]
This implies:
\[ 2B = A + C \]

### Step 2: Use the sum of angles in a triangle
We know that the sum of the angles in a triangle is \( 180^\circ \):
\[ A + B + C = 180^\circ \]
Substituting \( A + C = 2B \) into the equation:
\[ 2B + B = 180^\circ \]
This simplifies to:
\[ 3B = 180^\circ \]
Thus, we find:
\[ B = \frac{180^\circ}{3} = 60^\circ \]

### Step 3: Express angles A and C in terms of B
Since \( B = 60^\circ \), we can express \( A \) and \( C \) as:
\[ A = B - d = 60^\circ - d \]
\[ C = B + d = 60^\circ + d \]
where \( d \) is the common difference in the AP.

### Step 4: Use the cosine rule
We can apply the cosine rule to find relationships between the sides \( a, b, c \) of the triangle:
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]
Since \( B = 60^\circ \), we know:
\[ \cos 60^\circ = \frac{1}{2} \]
Substituting this into the cosine rule gives:
\[ \frac{1}{2} = \frac{a^2 + c^2 - b^2}{2ac} \]
Multiplying both sides by \( 2ac \):
\[ ac = a^2 + c^2 - b^2 \]
Rearranging this equation, we get:
\[ b^2 = a^2 + c^2 - ac \]

### Conclusion
Thus, the relationship we derived from the angles being in AP is:
\[ b^2 = a^2 + c^2 - ac \]