If in a triangle ABC `angle B=60^@` then

Options

A. `(a-b)^2 + ab =c^2`
B. `(b - c)^2+bc = a^2`
C. `(c-a)^2 + ca=b^2`
D. `a^2 + b^2 + c^2 = 2b^2 + ac`

Answer

To solve the problem, we will use the cosine rule in triangle ABC where angle B is given as 60 degrees.

### Step-by-Step Solution:

1. **Identify the Cosine Rule**:
The cosine rule states that for any triangle ABC, the following relationship holds:
\[ \cos B = \frac{A^2 + C^2 - B^2}{2AC} \]

2. **Substitute the Value of Angle B**:
Since angle B is given as 60 degrees, we can substitute this value into the cosine rule:
\[ \cos 60^\circ = \frac{A^2 + C^2 - B^2}{2AC} \]

3. **Calculate \(\cos 60^\circ\)**:
We know that:
\[ \cos 60^\circ = \frac{1}{2} \]
Therefore, we can rewrite the equation as:
\[ \frac{1}{2} = \frac{A^2 + C^2 - B^2}{2AC} \]

4. **Cross-Multiply to Eliminate the Fraction**:
Cross-multiplying gives us:
\[ 1 \cdot 2AC = A^2 + C^2 - B^2 \]
Simplifying this, we have:
\[ 2AC = A^2 + C^2 - B^2 \]

5. **Rearranging the Equation**:
Rearranging the equation to isolate \(B^2\) gives:
\[ B^2 = A^2 + C^2 - 2AC \]

6. **Recognizing the Perfect Square**:
The expression \(A^2 + C^2 - 2AC\) can be recognized as a perfect square:
\[ B^2 = (C - A)^2 \]

7. **Final Relation**:
Thus, we conclude that:
\[ B = |C - A| \]

### Summary:
The relationship derived from the triangle ABC with angle B equal to 60 degrees is:
\[ B = |C - A| \]