When any two sides and one of the opposite acute angle are given, under certain additional conditions two triangles are possible. The case when two triangles are possible is called the ambiguous case.
In fact when any two sides and the angle opposite to one of them are given either no triangle is posible or only one triangle is possible or two triangles are possible.
In the ambiguous case, let a,b and `angle A` are given and `c_(1), c_(2)` are two values of the third side c.
On the basis of above information, answer the following questions
Two different triangles are possible when

Options

A. `b sin A lt a`
B. `b sin A lt a and b gt a `
C. `b sin A lt a and b lt a `
D. `b sin A lt a and a =b`

Answer

To determine when two different triangles are possible given two sides and the angle opposite to one of them, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
We are given two sides, say \( a \) and \( b \), and an angle \( A \) opposite to side \( a \).

2. **Use the Law of Cosines**:
According to the Law of Cosines, we can express the relationship between the sides and the angle as follows:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
Rearranging this gives us:
\[ b^2 + c^2 - a^2 = 2bc \cos A \]

3. **Rearranging into Quadratic Form**:
We can rearrange the equation to form a quadratic equation in terms of \( c \):
\[ c^2 - 2b \cos A \cdot c + (b^2 - a^2) = 0 \]

4. **Identify the Quadratic Coefficients**:
In the standard form \( Ax^2 + Bx + C = 0 \), we have:
- \( A = 1 \)
- \( B = -2b \cos A \)
- \( C = b^2 - a^2 \)

5. **Apply the Discriminant Condition**:
For two distinct solutions (two triangles), the discriminant \( D \) of the quadratic must be greater than zero:
\[ D = B^2 - 4AC > 0 \]
Substituting the values:
\[ (-2b \cos A)^2 - 4 \cdot 1 \cdot (b^2 - a^2) > 0 \]
Simplifying this gives:
\[ 4b^2 \cos^2 A - 4(b^2 - a^2) > 0 \]
Dividing through by 4:
\[ b^2 \cos^2 A - (b^2 - a^2) > 0 \]

6. **Rearranging the Inequality**:
Rearranging the inequality yields:
\[ b^2 \cos^2 A > b^2 - a^2 \]
This can be rewritten as:
\[ a^2 > b^2(1 - \cos^2 A) \]
Using the identity \( \sin^2 A = 1 - \cos^2 A \):
\[ a^2 > b^2 \sin^2 A \]

7. **Final Conditions**:
Taking square roots gives us the condition:
\[ a > b \sin A \]
Additionally, since \( A \) is an angle in a triangle, we also have:
\[ b < a \]

### Conclusion:
Thus, two different triangles are possible when:
1. \( a > b \sin A \)
2. \( b < a \)