In a `triangleABC," if "(a)/(b^(2)-c^(2))+(c)/(b^(2)-a^(2))=0," then "angleB=`

Options

A. `pi/2`
B. `pi/4`
C. `(2pi)/(3)`
D. `pi/3`

Answer

To solve the problem, we need to find the angle \( B \) in triangle \( ABC \) given the equation:

\[ \frac{a}{b^2 - c^2} + \frac{c}{b^2 - a^2} = 0 \]

### Step-by-Step Solution:

1. **Rewrite the Equation**:
Start with the given equation:
\[ \frac{a}{b^2 - c^2} + \frac{c}{b^2 - a^2} = 0 \]
This implies:
\[ \frac{a}{b^2 - c^2} = -\frac{c}{b^2 - a^2} \]

2. **Cross Multiply**:
Cross multiplying gives:
\[ a(b^2 - a^2) = -c(b^2 - c^2) \]
Simplifying this, we have:
\[ ab^2 - a^3 = -cb^2 + c^3 \]

3. **Rearranging Terms**:
Rearranging the equation results in:
\[ ab^2 + cb^2 = a^3 + c^3 \]
Factoring out \( b^2 \) from the left side:
\[ (a + c)b^2 = a^3 + c^3 \]

4. **Using the Sum of Cubes Formula**:
Recall the identity for the sum of cubes:
\[ a^3 + c^3 = (a + c)(a^2 - ac + c^2) \]
Substituting this into our equation gives:
\[ (a + c)b^2 = (a + c)(a^2 - ac + c^2) \]

5. **Canceling Common Factors**:
If \( a + c \neq 0 \), we can divide both sides by \( a + c \):
\[ b^2 = a^2 - ac + c^2 \]

6. **Applying the Cosine Rule**:
According to the cosine rule in triangle \( ABC \):
\[ b^2 = a^2 + c^2 - 2ac \cos B \]
Setting the two expressions for \( b^2 \) equal gives:
\[ a^2 - ac + c^2 = a^2 + c^2 - 2ac \cos B \]

7. **Simplifying the Equation**:
Cancel \( a^2 \) and \( c^2 \) from both sides:
\[ -ac = -2ac \cos B \]
Dividing both sides by \( -ac \) (assuming \( ac \neq 0 \)):
\[ 1 = 2 \cos B \]

8. **Solving for \( \cos B \)**:
This simplifies to:
\[ \cos B = \frac{1}{2} \]

9. **Finding Angle \( B \)**:
The angle \( B \) for which \( \cos B = \frac{1}{2} \) is:
\[ B = 60^\circ \]
Therefore, the final answer is:
\[ \angle B = 60^\circ \]