In a triangle ABC

Options

A. sinA. sinB.sinC `=Delta/(2R^(2))`
B. sinA.sinB.sinC `=r/(2R)(sinA + sinB + sinC)`
C. `a cosA + b cosB + c cos C= (abc)/(2R^(2))`
D. `sinA. sinB.sinC = R/(2r)(sinA + sinB + sinC)`

Answer

To solve the problem regarding the triangle ABC and to verify the given equation, we will follow these steps:

### Step-by-Step Solution

1. **Identify the Given Information**:
We have a triangle ABC with sides a, b, and c opposite to angles A, B, and C respectively. We need to verify the equation:
\[ \sin A \cdot \sin B \cdot \sin C = \frac{\Delta}{2R^2} \]
where \(\Delta\) is the area of the triangle and \(R\) is the circumradius.

2. **Area of Triangle**:
The area \(\Delta\) of triangle ABC can be expressed using the formula:
\[ \Delta = \frac{1}{2}ab \sin C \]

3. **Using the Sine Rule**:
According to the sine rule:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \]
From this, we can express the sides in terms of the angles and the circumradius \(R\):
\[ a = 2R \sin A, \quad b = 2R \sin B, \quad c = 2R \sin C \]

4. **Substituting Values in Area Formula**:
Now, substituting the values of \(a\) and \(b\) into the area formula:
\[ \Delta = \frac{1}{2} (2R \sin A)(2R \sin B) \sin C \]
Simplifying this, we get:
\[ \Delta = 2R^2 \sin A \sin B \sin C \]

5. **Expressing \(\Delta\) in Terms of \(R\)**:
Now, we can express \(\Delta\) in terms of \(R\):
\[ \Delta = 2R^2 \sin A \sin B \sin C \]

6. **Finding \(\frac{\Delta}{2R^2}\)**:
To find \(\frac{\Delta}{2R^2}\):
\[ \frac{\Delta}{2R^2} = \sin A \sin B \sin C \]

7. **Conclusion**:
Thus, we have shown that:
\[ \sin A \cdot \sin B \cdot \sin C = \frac{\Delta}{2R^2} \]
Therefore, the equation is verified, and the first option is correct.