If in a triangle ABC ,`(b+c)/11=(c+a)/12=(a+b)/13` then cosA is equal to

Options

A. `1/5`
B. `5/7`
C. `19/35`
D. none of these

Answer

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

\[ \frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13} \]

Let's denote this common value as \( k \). Therefore, we can express \( b+c \), \( c+a \), and \( a+b \) in terms of \( k \):

1. \( b + c = 11k \)
2. \( c + a = 12k \)
3. \( a + b = 13k \)

Now, we can rearrange these equations to express \( a \), \( b \), and \( c \) in terms of \( k \):

From the first equation:
\[ c = 11k - b \tag{1} \]

From the second equation:
\[ a = 12k - c \tag{2} \]

From the third equation:
\[ b = 13k - a \tag{3} \]

Next, we can substitute equation (1) into equation (2):

Substituting \( c \) from (1) into (2):
\[ a = 12k - (11k - b) = 12k - 11k + b = k + b \tag{4} \]

Now, substitute equation (4) into equation (3):
\[ b = 13k - (k + b) \]
\[ b + b = 13k - k \]
\[ 2b = 12k \implies b = 6k \tag{5} \]

Now we can substitute \( b = 6k \) back into equation (4) to find \( a \):
\[ a = k + 6k = 7k \tag{6} \]

Next, substitute \( b = 6k \) into equation (1) to find \( c \):
\[ c = 11k - b = 11k - 6k = 5k \tag{7} \]

Now we have:
- \( a = 7k \)
- \( b = 6k \)
- \( c = 5k \)

Next, we need to find \( \cos A \) using the cosine rule:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]

Substituting the values of \( a \), \( b \), and \( c \):
\[ \cos A = \frac{(6k)^2 + (5k)^2 - (7k)^2}{2 \cdot 6k \cdot 5k} \]
\[ = \frac{36k^2 + 25k^2 - 49k^2}{60k^2} \]
\[ = \frac{(36 + 25 - 49)k^2}{60k^2} \]
\[ = \frac{12k^2}{60k^2} \]
\[ = \frac{12}{60} = \frac{1}{5} \]

Thus, the value of \( \cos A \) is:
\[ \cos A = \frac{1}{5} \]