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.

Answer

To solve the problem, we need to find the value of sin A in triangle ABC given the sides a, c, and angle A, along with two values of the third side b (b1 and b2), where b2 = 2b1.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- We have a triangle ABC with sides a, b, and c.
- We know angle A and the sides a and c.
- We have two values for side b: b1 and b2, where b2 = 2b1.

2. **Using the Cosine Rule**:
The cosine rule states:
\[ b^2 = a^2 + c^2 - 2ac \cdot \cos A \]
We can apply this for both b1 and b2:
- For b1:
\[ b1^2 = a^2 + c^2 - 2ac \cdot \cos A \quad (1) \]
- For b2:
\[ b2^2 = a^2 + c^2 - 2ac \cdot \cos A \quad (2) \]

3. **Substituting b2 = 2b1**:
From the second equation (2), substituting b2:
\[ (2b1)^2 = a^2 + c^2 - 2ac \cdot \cos A \]
This simplifies to:
\[ 4b1^2 = a^2 + c^2 - 2ac \cdot \cos A \quad (3) \]

4. **Setting Up the Equations**:
Now we have two equations:
- From (1):
\[ b1^2 = a^2 + c^2 - 2ac \cdot \cos A \]
- From (3):
\[ 4b1^2 = a^2 + c^2 - 2ac \cdot \cos A \]

5. **Equating the Two Expressions**:
From (1) and (3):
\[ 4(a^2 + c^2 - 2ac \cdot \cos A) = a^2 + c^2 - 2ac \cdot \cos A \]
Rearranging gives:
\[ 4b1^2 = 3(a^2 + c^2 - 2ac \cdot \cos A) \]

6. **Finding the Value of cos A**:
Rearranging the equation:
\[ 3b1^2 = a^2 + c^2 - 2ac \cdot \cos A \]
This implies:
\[ 2ac \cdot \cos A = a^2 + c^2 - 3b1^2 \]
Thus:
\[ \cos A = \frac{a^2 + c^2 - 3b1^2}{2ac} \]

7. **Finding sin A**:
Using the identity \( \sin^2 A + \cos^2 A = 1 \):
\[ \sin^2 A = 1 - \cos^2 A \]
Substitute \( \cos A \):
\[ \sin^2 A = 1 - \left( \frac{a^2 + c^2 - 3b1^2}{2ac} \right)^2 \]

8. **Final Expression for sin A**:
Taking the square root gives:
\[ \sin A = \sqrt{1 - \left( \frac{a^2 + c^2 - 3b1^2}{2ac} \right)^2} \]