In `DeltaABC, /_A` is right angle and AD is perpendicular to BC . If `AD=4cm, BC=12` cm, then value of `(cot B+cot C)` is

Options

A. `6`
B. `3`
C. `4`
D. `(3)/(2)`

Answer

To solve the problem, we need to find the value of \( \cot B + \cot C \) in triangle \( \Delta ABC \) where \( \angle A \) is a right angle, \( AD \) is perpendicular to \( BC \), \( AD = 4 \, \text{cm} \), and \( BC = 12 \, \text{cm} \).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- \( AD = 4 \, \text{cm} \)
- \( BC = 12 \, \text{cm} \)

2. **Use the Area of Triangle:**
The area of triangle \( ABC \) can be calculated using the base \( BC \) and height \( AD \):
\[ \text{Area} = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times 12 \times 4 = 24 \, \text{cm}^2 \]

3. **Relate the Area to Sides \( AB \) and \( AC \):**
The area can also be expressed in terms of sides \( AB \) and \( AC \) as:
\[ \text{Area} = \frac{1}{2} \times AB \times AC \]
Therefore, we have:
\[ AB \times AC = 48 \]

4. **Express \( AB \) and \( AC \) in Terms of Angles:**
Using the definitions of sine:
\[ AB = \frac{AD}{\sin B} = \frac{4}{\sin B} \]
\[ AC = \frac{AD}{\sin C} = \frac{4}{\sin C} \]
Now substitute these into the area equation:
\[ \left(\frac{4}{\sin B}\right) \left(\frac{4}{\sin C}\right) = 48 \]
Simplifying gives:
\[ \frac{16}{\sin B \sin C} = 48 \]
Thus:
\[ \sin B \sin C = \frac{16}{48} = \frac{1}{3} \]

5. **Find \( \cot B + \cot C \):**
We know:
\[ \cot B = \frac{\cos B}{\sin B}, \quad \cot C = \frac{\cos C}{\sin C} \]
Therefore:
\[ \cot B + \cot C = \frac{\cos B}{\sin B} + \frac{\cos C}{\sin C} = \frac{\cos B \sin C + \cos C \sin B}{\sin B \sin C} \]
Using the identity \( \sin(B + C) = \sin(90^\circ) = 1 \):
\[ \sin B \cos C + \cos B \sin C = \sin(B + C) = 1 \]
Thus:
\[ \cot B + \cot C = \frac{1}{\sin B \sin C} \]

6. **Substitute the Value of \( \sin B \sin C \):**
\[ \cot B + \cot C = \frac{1}{\frac{1}{3}} = 3 \]

### Final Answer:
The value of \( \cot B + \cot C \) is \( 3 \).