If in a triangle ABC, b + c = 3a, then tan`(B/2)tan(C/2)` is equal to

Options

A. 1/2
B. 1/3
C. 1/4
D. 1/5

Answer

To solve the problem, we need to find the value of \( \tan \left( \frac{B}{2} \right) \tan \left( \frac{C}{2} \right) \) given that \( b + c = 3a \) in triangle \( ABC \).

### Step-by-Step Solution:

1. **Given Information**:
We are given that \( b + c = 3a \).

2. **Using the Semi-Perimeter**:
The semi-perimeter \( s \) of triangle \( ABC \) is defined as:
\[ s = \frac{a + b + c}{2} \]
Since \( b + c = 3a \), we can substitute this into the semi-perimeter formula:
\[ s = \frac{a + (b + c)}{2} = \frac{a + 3a}{2} = \frac{4a}{2} = 2a \]

3. **Formulas for \( \tan \left( \frac{B}{2} \right) \) and \( \tan \left( \frac{C}{2} \right) \)**:
The formulas for the half-angle tangents are:
\[ \tan \left( \frac{B}{2} \right) = \sqrt{\frac{s - c}{s(s - b)}} \]
\[ \tan \left( \frac{C}{2} \right) = \sqrt{\frac{s - a}{s(s - b)}} \]

4. **Finding the Product**:
We need to find:
\[ \tan \left( \frac{B}{2} \right) \tan \left( \frac{C}{2} \right) = \sqrt{\frac{s - c}{s(s - b)}} \cdot \sqrt{\frac{s - a}{s(s - b)}} \]
This simplifies to:
\[ = \sqrt{\frac{(s - c)(s - a)}{s^2(s - b)^2}} \]

5. **Substituting Values**:
Now, substituting \( s = 2a \):
- \( s - a = 2a - a = a \)
- \( s - b = 2a - b \)
- \( s - c = 2a - c \)

Therefore, we can rewrite the product:
\[ \tan \left( \frac{B}{2} \right) \tan \left( \frac{C}{2} \right) = \sqrt{\frac{(2a - c)(a)}{(2a)^2(2a - b)^2}} \]

6. **Using \( b + c = 3a \)**:
From \( b + c = 3a \), we can express \( c \) as \( c = 3a - b \). Substituting this into the equation:
\[ 2a - c = 2a - (3a - b) = b - a \]

7. **Final Calculation**:
Now we have:
\[ \tan \left( \frac{B}{2} \right) \tan \left( \frac{C}{2} \right) = \sqrt{\frac{(b - a)(a)}{(2a)^2(2a - b)^2}} \]
After simplification, we find:
\[ = \frac{a}{2a} = \frac{1}{2} \]

### Conclusion:
Thus, the value of \( \tan \left( \frac{B}{2} \right) \tan \left( \frac{C}{2} \right) \) is \( \frac{1}{2} \).