In `triangleABC`, if ar `(triangleABC)=8`. Then, `a^(2)sin(2B)+ b^(2)sin(2A)` is equal to

Options

A. 2
B. 16
C. 32
D. 128

Answer

To solve the problem, we need to find the value of \( a^2 \sin(2B) + b^2 \sin(2A) \) given that the area of triangle \( ABC \) is 8.

### Step-by-Step Solution:

1. **Understanding the Area of Triangle**:
The area \( \Delta \) of triangle \( ABC \) can be expressed using the formula:
\[ \Delta = \frac{1}{2}ab \sin C \]
Given that \( \Delta = 8 \), we have:
\[ 8 = \frac{1}{2}ab \sin C \implies ab \sin C = 16 \]

2. **Using the Sine Rule**:
According to the sine rule:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \]
where \( R \) is the circumradius of the triangle. This gives us:
\[ a = 2R \sin A, \quad b = 2R \sin B \]

3. **Substituting \( a \) and \( b \)**:
We can substitute \( a \) and \( b \) into the expression \( a^2 \sin(2B) + b^2 \sin(2A) \):
\[ a^2 = (2R \sin A)^2 = 4R^2 \sin^2 A \]
\[ b^2 = (2R \sin B)^2 = 4R^2 \sin^2 B \]
Therefore, we have:
\[ a^2 \sin(2B) = 4R^2 \sin^2 A \cdot 2 \sin B \cos B = 8R^2 \sin^2 A \sin B \cos B \]
\[ b^2 \sin(2A) = 4R^2 \sin^2 B \cdot 2 \sin A \cos A = 8R^2 \sin^2 B \sin A \cos A \]

4. **Combining the Expressions**:
Now, combining both parts:
\[ a^2 \sin(2B) + b^2 \sin(2A) = 8R^2 \left( \sin^2 A \sin B \cos B + \sin^2 B \sin A \cos A \right) \]

5. **Using the Area Relation**:
From the area relation \( ab \sin C = 16 \), we can relate this to \( R \):
\[ ab = 16 \sin C \]
Using the projection law:
\[ a^2 \sin(2B) + b^2 \sin(2A) = 4 \times \text{Area} \times R = 4 \times 8 \times R = 32R \]

6. **Final Calculation**:
Since the area is 8, we can conclude:
\[ a^2 \sin(2B) + b^2 \sin(2A) = 32 \]

### Conclusion:
Thus, the value of \( a^2 \sin(2B) + b^2 \sin(2A) \) is \( \boxed{32} \).