In triangle, ABC if `2a^(2) b^(2) + 2b^(2) c^(2) = a^(4) + b^(4) + c^(4)`, then angle B is equal to

Options

A. `45^(@)`
B. `135^(@)`
C. `120^(@)`
D. `60^(@)`

Explanation

To solve the problem, we start with the given equation:

### Step 1: Write down the equation
We have:
\[ 2a^2b^2 + 2b^2c^2 = a^4 + b^4 + c^4 \]

### Step 2: Rearranging the equation
We can rearrange the equation to isolate the terms involving squares:
\[ a^4 + b^4 + c^4 - 2a^2b^2 - 2b^2c^2 = 0 \]

### Step 3: Use the identity
We can use the identity:
\[ a^4 + b^4 + c^4 - 2a^2b^2 - 2b^2c^2 - 2c^2a^2 = (a^2 - b^2 + c^2)^2 \]
This implies:
\[ (a^2 - b^2 + c^2)^2 = 2(a^2c^2) \]

### Step 4: Taking square roots
Taking the square root of both sides gives us:
\[ a^2 - b^2 + c^2 = \pm \sqrt{2}ac \]

### Step 5: Rearranging to find cos B
Dividing both sides by \( 2ac \):
\[ \frac{a^2 - b^2 + c^2}{2ac} = \pm \frac{1}{\sqrt{2}} \]

### Step 6: Using the cosine rule
From the cosine rule, we know:
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]
Thus, we can equate:
\[ \cos B = \pm \frac{1}{\sqrt{2}} \]

### Step 7: Finding angle B
The values of \( \cos B = \frac{1}{\sqrt{2}} \) correspond to:
\[ B = 45^\circ \]
And \( \cos B = -\frac{1}{\sqrt{2}} \) corresponds to:
\[ B = 135^\circ \]

### Conclusion
Thus, the possible values for angle B are:
\[ B = 45^\circ \text{ or } 135^\circ \]