Given `(b+c)/(11) = (c+a)/(12) = (a+b)/(13)` for a `Delta ABC` with usual notation. If `(cos A)/(alpha) = (cos B)/(beta) = (cos C)/(gamma)`, then the ordered triad `(alpha, beta, gamma)` has a value:

Options

A. (3, 4, 5)
B. (5, 12, 13)
C. (7, 19, 25)
D. (19, 7, 25)

Answer

To solve the problem step by step, we will start from the given equations and derive the values of \( a \), \( b \), and \( c \) in terms of \( \lambda \), and then find the ordered triad \( (\alpha, \beta, \gamma) \).

### Step 1: Set up the equations
Given:
\[ \frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13} = \lambda \]
From this, we can write three equations:
1. \( b + c = 11\lambda \)
2. \( c + a = 12\lambda \)
3. \( a + b = 13\lambda \)

### Step 2: Express \( a \), \( b \), and \( c \) in terms of \( \lambda \)
From the first equation:
\[ c = 11\lambda - b \tag{1} \]

Substituting (1) into the second equation:
\[ (11\lambda - b) + a = 12\lambda \implies a = 12\lambda - 11\lambda + b \implies a = \lambda + b \tag{2} \]

Now substituting (2) into the third equation:
\[ (\lambda + b) + b = 13\lambda \implies \lambda + 2b = 13\lambda \implies 2b = 12\lambda \implies b = 6\lambda \tag{3} \]

### Step 3: Find \( a \) and \( c \)
Using (3) in (2):
\[ a = \lambda + 6\lambda = 7\lambda \tag{4} \]

Using (3) in (1):
\[ c = 11\lambda - 6\lambda = 5\lambda \tag{5} \]

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

### Step 4: Calculate \( \cos A \), \( \cos B \), and \( \cos C \)
Using the cosine rule:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
Substituting values:
\[ \cos A = \frac{(6\lambda)^2 + (5\lambda)^2 - (7\lambda)^2}{2 \cdot 6\lambda \cdot 5\lambda} = \frac{36\lambda^2 + 25\lambda^2 - 49\lambda^2}{60\lambda^2} = \frac{12\lambda^2}{60\lambda^2} = \frac{1}{5} \]

Now for \( \cos B \):
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]
Substituting values:
\[ \cos B = \frac{(7\lambda)^2 + (5\lambda)^2 - (6\lambda)^2}{2 \cdot 7\lambda \cdot 5\lambda} = \frac{49\lambda^2 + 25\lambda^2 - 36\lambda^2}{70\lambda^2} = \frac{38\lambda^2}{70\lambda^2} = \frac{19}{35} \]

Now for \( \cos C \):
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]
Substituting values:
\[ \cos C = \frac{(7\lambda)^2 + (6\lambda)^2 - (5\lambda)^2}{2 \cdot 7\lambda \cdot 6\lambda} = \frac{49\lambda^2 + 36\lambda^2 - 25\lambda^2}{84\lambda^2} = \frac{60\lambda^2}{84\lambda^2} = \frac{5}{7} \]

### Step 5: Find \( \alpha, \beta, \gamma \)
From the problem statement:
\[ \frac{\cos A}{\alpha} = \frac{\cos B}{\beta} = \frac{\cos C}{\gamma} = k \]
Thus:
\[ \alpha = \frac{\cos A}{k}, \quad \beta = \frac{\cos B}{k}, \quad \gamma = \frac{\cos C}{k} \]
Let \( k = \frac{1}{5\alpha} \), then:
\[ \frac{1}{5} = k\alpha \implies \alpha = 5k \]
\[ \frac{19}{35} = k\beta \implies \beta = \frac{19}{35k} \]
\[ \frac{5}{7} = k\gamma \implies \gamma = \frac{5}{7k} \]

### Step 6: Find the ordered triad \( (\alpha, \beta, \gamma) \)
Setting \( k = 1 \):
\[ \alpha = 5, \quad \beta = \frac{19}{35}, \quad \gamma = \frac{5}{7} \]

To find the ordered triad:
\[ \alpha : \beta : \gamma = 5 : \frac{19}{35} : \frac{5}{7} \]
Multiplying through by \( 35 \):
\[ \alpha : \beta : \gamma = 175 : 19 : 25 \]

Thus, the ordered triad \( (\alpha, \beta, \gamma) \) is:
\[ \boxed{(7, 19, 25)} \]