In a triangle ABC , if `(b+c)/(11)=(c+a)/(12)=(a+b)/(13)`, then cos A=

Options

A. (a)`(19)/(35)`
B. (b)`(5)/(7)`
C. (c)`(1)/(5)`
D. (d)`(11)/(12)`

Answer

To solve the problem step by step, we start with the given equations:

1. **Given equations**:
\[ \frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13} = \lambda \]

2. **Express \(b+c\), \(c+a\), and \(a+b\) in terms of \(\lambda\)**:
\[ b+c = 11\lambda \quad \text{(1)} \]
\[ c+a = 12\lambda \quad \text{(2)} \]
\[ a+b = 13\lambda \quad \text{(3)} \]

3. **Add equations (1), (2), and (3)**:
\[ (b+c) + (c+a) + (a+b) = 11\lambda + 12\lambda + 13\lambda \]
This simplifies to:
\[ 2(a+b+c) = 36\lambda \]
Therefore, we can express \(a+b+c\) as:
\[ a+b+c = 18\lambda \quad \text{(4)} \]

4. **Now, we can express \(a\), \(b\), and \(c\) in terms of \(\lambda\)**:
- From equation (1):
\[ a = (b+c) - (b+c) + a = 18\lambda - 11\lambda = 7\lambda \]
- From equation (2):
\[ b = (c+a) - (c+a) + b = 18\lambda - 12\lambda = 6\lambda \]
- From equation (3):
\[ c = (a+b) - (a+b) + c = 18\lambda - 13\lambda = 5\lambda \]

5. **Now we have**:
\[ a = 7\lambda, \quad b = 6\lambda, \quad c = 5\lambda \]

6. **Use the cosine rule to find \(\cos A\)**:
The cosine rule states:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]

7. **Substituting the values of \(a\), \(b\), and \(c\)**:
\[ b^2 = (6\lambda)^2 = 36\lambda^2 \]
\[ c^2 = (5\lambda)^2 = 25\lambda^2 \]
\[ a^2 = (7\lambda)^2 = 49\lambda^2 \]

8. **Substituting into the cosine formula**:
\[ \cos A = \frac{36\lambda^2 + 25\lambda^2 - 49\lambda^2}{2 \cdot (6\lambda)(5\lambda)} \]
Simplifying the numerator:
\[ \cos A = \frac{36\lambda^2 + 25\lambda^2 - 49\lambda^2}{60\lambda^2} \]
\[ = \frac{12\lambda^2}{60\lambda^2} \]
\[ = \frac{12}{60} = \frac{1}{5} \]

9. **Final result**:
\[ \cos A = \frac{1}{5} \]