If in `Delta ABC, b = 3 cm, c = 4 cm` and the length of the perpendicular from A to the side BC is 2 cm, then how many such triangle are possible ?

Explanation

To determine how many triangles can be formed with the given conditions, we will follow these steps:

### Step-by-Step Solution:

1. **Identify Given Information**:
- Side \( b = 3 \, \text{cm} \)
- Side \( c = 4 \, \text{cm} \)
- Length of the perpendicular from \( A \) to side \( BC \) (denote as \( AD \)) is \( 2 \, \text{cm} \).

2. **Draw the Triangle**:
- Draw triangle \( ABC \) with \( B \) and \( C \) as the base and \( A \) above the base such that \( AD \) is perpendicular to \( BC \).

3. **Use the Area Formula**:
- The area \( A \) of triangle \( ABC \) can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BC \times AD \]
- Here, \( AD = 2 \, \text{cm} \), so:
\[ A = \frac{1}{2} \times BC \times 2 = BC \]

4. **Apply the Law of Sines**:
- According to the Law of Sines:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
- We can express \( BC \) in terms of \( b \) and \( c \) using the sine of angles \( B \) and \( C \).

5. **Find \( \sin B \)**:
- From the triangle, we know that:
\[ AD = c \sin B \implies 2 = 4 \sin B \implies \sin B = \frac{1}{2} \]
- This implies that \( B \) can be \( 30^\circ \) or \( 150^\circ \).

6. **Calculate \( \cos B \)**:
- Using the identity \( \cos^2 B + \sin^2 B = 1 \):
\[ \cos^2 B = 1 - \left(\frac{1}{2}\right)^2 = 1 - \frac{1}{4} = \frac{3}{4} \implies \cos B = \frac{\sqrt{3}}{2} \]

7. **Use the Cosine Rule**:
- Applying the cosine rule:
\[ c^2 = a^2 + b^2 - 2ab \cos B \]
- Substitute \( b = 3 \), \( c = 4 \), and \( \cos B = \frac{\sqrt{3}}{2} \):
\[ 16 = a^2 + 9 - 2 \cdot 3 \cdot a \cdot \frac{\sqrt{3}}{2} \]
- Simplifying gives:
\[ 16 = a^2 + 9 - 3\sqrt{3}a \]
\[ a^2 - 3\sqrt{3}a - 7 = 0 \]

8. **Calculate the Discriminant**:
- The discriminant \( D \) of the quadratic equation \( a^2 - 3\sqrt{3}a - 7 = 0 \) is:
\[ D = b^2 - 4ac = (3\sqrt{3})^2 - 4 \cdot 1 \cdot (-7) = 27 + 28 = 55 \]

9. **Determine the Nature of Roots**:
- Since \( D > 0 \), there are two distinct real roots for \( a \).

10. **Conclusion**:
- Therefore, there are **two possible triangles** that can be formed with the given conditions.