In a triangle ABC,vertex angles A,B,C and side BC are given .The area of `triangle ABC` is

Options

A. `(s(s-a)(s-b)(s-c))/2`
B. `(b^(2)sinCsinA)/(sinB)`
C. ab sin C
D. `1/2(a^(2)sinBsinC)/(sinA)`

Answer

To find the area of triangle ABC given the vertex angles A, B, C and side BC (denoted as A), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information**:
- We have a triangle ABC with angles A, B, C.
- The side opposite to angle A (BC) is given as A.

2. **Use the Formula for the Area of a Triangle**:
- The area \( \text{Area} \) of triangle ABC can be expressed using the formula:
\[ \text{Area} = \frac{1}{2} \times A \times B \times \sin C \]
- Here, \( A \) is the length of side BC, \( B \) is the length of side AC, and \( C \) is the angle at vertex A.

3. **Apply the Sine Rule**:
- According to the sine rule, we have:
\[ \frac{A}{\sin A} = \frac{B}{\sin B} \]
- From this, we can express \( B \) in terms of \( A \) and \( \sin A \):
\[ B = \frac{A \cdot \sin B}{\sin A} \]

4. **Substitute B into the Area Formula**:
- Now substitute the expression for \( B \) into the area formula:
\[ \text{Area} = \frac{1}{2} \times A \times \left(\frac{A \cdot \sin B}{\sin A}\right) \times \sin C \]

5. **Simplify the Expression**:
- Simplifying the area expression gives:
\[ \text{Area} = \frac{1}{2} \times A^2 \times \frac{\sin B \cdot \sin C}{\sin A} \]

6. **Final Result**:
- Therefore, the area of triangle ABC is:
\[ \text{Area} = \frac{1}{2} A^2 \frac{\sin B \cdot \sin C}{\sin A} \]

### Conclusion:
The area of triangle ABC, given the angles A, B, C and side BC, is:
\[ \text{Area} = \frac{1}{2} A^2 \frac{\sin B \cdot \sin C}{\sin A} \]