In a `delta ABC,` a,c, A are given and `b_(1) , b_(2)` are two values of third side b such that `b_(2)=2b_(1).` Then, the value of sin A.

Options

A. `sqrt((9a^(2)-c^(2))/(8a^(2)))`
B. `sqrt((9a^(2)-c ^(2))/(8c^(2)))`
C. `sqrt((9a^(2)-c ^(2))/(8b^(2)))`
D. None of these

Answer

To solve the problem step by step, we will use the given information about triangle \(ABC\) and the relationship between the sides and angles.

### Step 1: Understand the given information
We have a triangle \(ABC\) with sides \(a\), \(b\), \(c\) and angle \(A\) given. We know that there are two values for the side \(b\), namely \(b_1\) and \(b_2\), where \(b_2 = 2b_1\).

### Step 2: Use the Law of Cosines
According to the Law of Cosines, we can relate the sides and angles of the triangle as follows:
\[ b^2 = a^2 + c^2 - 2ac \cos A \]
This equation can be rearranged to form a quadratic equation in terms of \(b\):
\[ b^2 - 2ac \cos A + (c^2 - a^2) = 0 \]

### Step 3: Identify the coefficients
In the quadratic equation \(b^2 - 2ac \cos A + (c^2 - a^2) = 0\):
- Coefficient of \(b\) (linear term) is \(-2ac \cos A\)
- Constant term is \(c^2 - a^2\)

### Step 4: Sum and Product of Roots
For a quadratic equation \(Ax^2 + Bx + C = 0\), the sum of the roots \(b_1 + b_2 = -\frac{B}{A}\) and the product of the roots \(b_1 b_2 = \frac{C}{A}\).

Here, we have:
- Sum of the roots: \(b_1 + b_2 = 2ac \cos A\)
- Product of the roots: \(b_1 b_2 = c^2 - a^2\)

### Step 5: Substitute \(b_2 = 2b_1\)
Substituting \(b_2 = 2b_1\) into the sum of the roots:
\[ b_1 + 2b_1 = 3b_1 = 2ac \cos A \]
From this, we can express \(b_1\):
\[ b_1 = \frac{2ac \cos A}{3} \]

### Step 6: Substitute into the product of the roots
Now substituting \(b_1\) and \(b_2\) into the product of the roots:
\[ b_1 \cdot (2b_1) = 2b_1^2 = c^2 - a^2 \]
Substituting \(b_1 = \frac{2ac \cos A}{3}\):
\[ 2\left(\frac{2ac \cos A}{3}\right)^2 = c^2 - a^2 \]
This simplifies to:
\[ \frac{8a^2c^2 \cos^2 A}{9} = c^2 - a^2 \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ c^2 - \frac{8a^2c^2 \cos^2 A}{9} = a^2 \]
Multiplying through by 9 to eliminate the fraction:
\[ 9c^2 - 8a^2c^2 \cos^2 A = 9a^2 \]

### Step 8: Isolate \(\sin^2 A\)
Now we can express \(\sin^2 A\) in terms of \(a\) and \(c\):
\[ 8a^2c^2 \cos^2 A = 9c^2 - 9a^2 \]
Using the identity \(\sin^2 A + \cos^2 A = 1\):
\[ \cos^2 A = 1 - \sin^2 A \]
Substituting this into our equation:
\[ 8a^2c^2(1 - \sin^2 A) = 9c^2 - 9a^2 \]
Expanding and rearranging gives:
\[ 8a^2c^2 - 8a^2c^2 \sin^2 A = 9c^2 - 9a^2 \]
\[ 8a^2c^2 \sin^2 A = 9a^2 - 9c^2 + 8a^2c^2 \]

### Step 9: Solve for \(\sin^2 A\)
Finally, we can isolate \(\sin^2 A\):
\[ \sin^2 A = \frac{9a^2 - 9c^2 + 8a^2c^2}{8a^2c^2} \]

### Step 10: Conclusion
Thus, we find the value of \(\sin A\):
\[ \sin A = \sqrt{\frac{9a^2 - c^2}{8c^2}} \]

### Final Answer
The value of \(\sin A\) is:
\[ \sin A = \sqrt{\frac{9a^2 - c^2}{8c^2}} \]