In triangle ABC,a= 5,b=7 and sin A =3/4 how many such triangle are possible

Options

A. 1
B. 0
C. 2
D. Infinite

Answer

To determine how many triangles are possible given the values \( a = 5 \), \( b = 7 \), and \( \sin A = \frac{3}{4} \), we can use the Law of Sines.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- \( a = 5 \)
- \( b = 7 \)
- \( \sin A = \frac{3}{4} \)

2. **Apply the Law of Sines:**
The Law of Sines states that:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} \]
Substituting the known values:
\[ \frac{5}{\frac{3}{4}} = \frac{7}{\sin B} \]

3. **Simplify the Equation:**
First, simplify \( \frac{5}{\frac{3}{4}} \):
\[ \frac{5 \times 4}{3} = \frac{20}{3} \]
Now, we can rewrite the equation:
\[ \frac{20}{3} = \frac{7}{\sin B} \]

4. **Cross-Multiply to Solve for \( \sin B \):**
Cross-multiplying gives:
\[ 20 \sin B = 21 \]
Therefore, solving for \( \sin B \):
\[ \sin B = \frac{21}{20} \]

5. **Analyze the Value of \( \sin B \):**
The value \( \sin B = \frac{21}{20} \) is greater than 1. Since the sine of an angle cannot exceed 1, this indicates that there is no possible angle \( B \) that satisfies this condition.

6. **Conclusion:**
Since we found that \( \sin B \) exceeds the maximum possible value for sine, we conclude that **no triangle is possible** with the given conditions.

### Final Answer:
**No triangle is possible.**