In `triangleABC, (sinB)/(sin(A+B))` =

Options

A. `(b)/(a+b)`
B. `(b)/(c)`
C. `(-b)/(c)`
D. `(c)/(b)`

Answer

To solve the problem, we need to find the value of \(\frac{\sin B}{\sin(A+B)}\) in triangle \(ABC\).

### Step-by-Step Solution:

1. **Understanding the Angles in Triangle**:
In triangle \(ABC\), the sum of angles is given by:
\[ A + B + C = 180^\circ \]
Therefore, we can express \(A + B\) as:
\[ A + B = 180^\circ - C \]

2. **Using the Sine Function**:
We know that:
\[ \sin(A + B) = \sin(180^\circ - C) \]
Using the sine identity, we have:
\[ \sin(180^\circ - C) = \sin C \]
Thus, we can rewrite \(\sin(A + B)\):
\[ \sin(A + B) = \sin C \]

3. **Substituting Back into the Original Expression**:
Now, substituting this back into our original expression:
\[ \frac{\sin B}{\sin(A + B)} = \frac{\sin B}{\sin C} \]

4. **Applying the Sine Rule**:
According to the sine rule in triangle \(ABC\):
\[ \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} \]
From this, we can express \(\frac{\sin B}{\sin C}\):
\[ \frac{\sin B}{\sin C} = \frac{b}{c} \]

5. **Final Result**:
Therefore, we can conclude that:
\[ \frac{\sin B}{\sin(A + B)} = \frac{b}{c} \]

### Conclusion:
The value of \(\frac{\sin B}{\sin(A + B)}\) is \(\frac{b}{c}\), which corresponds to the second option provided in the question.