With usual notations, if in a triangle ABC `(b+c)/(11) = (c+a)/(12) = (a+b)/(13)`, then cos A : cos B : cos C is :

Options

A. `7:19,25`
B. `19:7:25`
C. `12:14:20`
D. `19:25:20`

Answer

To solve the problem, we start with the equations given in the triangle ABC:

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

Let this common ratio be \( k \). Then we can express \( b+c \), \( c+a \), and \( a+b \) in terms of \( k \):

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

Next, we can add all three equations:

\[ (b+c) + (c+a) + (a+b) = 11k + 12k + 13k \]

This simplifies to:

\[ 2a + 2b + 2c = 36k \]

Dividing through by 2 gives:

\[ a + b + c = 18k \]

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

From \( b + c = 11k \):

\[ a = (a + b + c) - (b + c) = 18k - 11k = 7k \]

From \( c + a = 12k \):

\[ b = (a + b + c) - (c + a) = 18k - 12k = 6k \]

From \( a + b = 13k \):

\[ c = (a + b + c) - (a + b) = 18k - 13k = 5k \]

Now we have:

- \( a = 7k \)
- \( b = 6k \)
- \( c = 5k \)

Next, we will use the cosine rule to find \( \cos A \), \( \cos B \), and \( \cos C \).

Using the cosine rule:

1. For \( \cos A \):

\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]

Substituting the values:

\[ \cos A = \frac{(6k)^2 + (5k)^2 - (7k)^2}{2 \cdot (6k)(5k)} = \frac{36k^2 + 25k^2 - 49k^2}{60k^2} = \frac{12k^2}{60k^2} = \frac{1}{5} \]

2. For \( \cos B \):

\[ \cos B = \frac{c^2 + a^2 - b^2}{2ca} \]

Substituting the values:

\[ \cos B = \frac{(5k)^2 + (7k)^2 - (6k)^2}{2 \cdot (5k)(7k)} = \frac{25k^2 + 49k^2 - 36k^2}{70k^2} = \frac{38k^2}{70k^2} = \frac{19}{35} \]

3. For \( \cos C \):

\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]

Substituting the values:

\[ \cos C = \frac{(7k)^2 + (6k)^2 - (5k)^2}{2 \cdot (7k)(6k)} = \frac{49k^2 + 36k^2 - 25k^2}{84k^2} = \frac{60k^2}{84k^2} = \frac{5}{7} \]

Now we have:

- \( \cos A = \frac{1}{5} \)
- \( \cos B = \frac{19}{35} \)
- \( \cos C = \frac{5}{7} \)

Finally, we need to express \( \cos A : \cos B : \cos C \):

\[ \cos A : \cos B : \cos C = \frac{1}{5} : \frac{19}{35} : \frac{5}{7} \]

To eliminate the fractions, we can find a common denominator, which is 35:

\[ \cos A : \cos B : \cos C = \frac{7}{35} : \frac{19}{35} : \frac{25}{35} = 7 : 19 : 25 \]

Thus, the final answer is:

\[ \cos A : \cos B : \cos C = 7 : 19 : 25 \]