Question

In a triangle \( ABC \), with usual notation, if \[ \frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13} \] then the ratio \( \cos A : \cos B : \cos C \) is:

• A. \( 19 : 7 : 25 \)
• B. \( 7 : 19 : 25 \)
• C. \( 12 : 14 : 20 \)
• D. \( 19 : 25 : 7 \)

Hint:

To find cosine ratios in a triangle when you have conditions involving sums of sides, express all sides in terms of a common variable, then apply the Law of Cosines.

Answer 1

The Correct Option is B

We are given: \[ \frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13} \] Let that common ratio be \( k \). Then: \[ b + c = 11k \quad \text{(1)}
c + a = 12k \quad \text{(2)}
a + b = 13k \quad \text{(3)} \] Step 1: Solve these equations to find \( a, b, c \) in terms of \( k \) Add (1) and (2): \[ b + c + c + a = 11k + 12k \Rightarrow a + b + 2c = 23k \quad \text{(4)} \] Subtract (3): \[ (4) - (3): (a + b + 2c) - (a + b) = 23k - 13k \Rightarrow 2c = 10k \Rightarrow c = 5k \] Substitute \( c = 5k \) into (1): \[ b + 5k = 11k \Rightarrow b = 6k \] Substitute \( c = 5k \) into (2): \[ a + 5k = 12k \Rightarrow a = 7k \] So we have: \[ a = 7k, \quad b = 6k, \quad c = 5k \] Step 2: Use Law of Cosines From the Law of Cosines: \[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{(6k)^2 + (5k)^2 - (7k)^2}{2 \cdot 6k \cdot 5k} = \frac{36k^2 + 25k^2 - 49k^2}{60k^2} = \frac{12k^2}{60k^2} = \frac{1}{5} \] \[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{49k^2 + 25k^2 - 36k^2}{2 \cdot 7k \cdot 5k} = \frac{38k^2}{70k^2} = \frac{19}{35} \] \[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{49k^2 + 36k^2 - 25k^2}{2 \cdot 7k \cdot 6k} = \frac{60k^2}{84k^2} = \frac{5}{7} \] Now express all ratios with common denominator: - \( \cos A = \frac{1}{5} = \frac{7}{35} \) - \( \cos B = \frac{19}{35} \) - \( \cos C = \frac{25}{35} \) Thus: \[ \cos A : \cos B : \cos C = 7 : 19 : 25 \] \[ \boxed{\text{Correct answer is Option (2)}} \]