Question

With usual notation, in a triangle ABC $\frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13}$, then the value of $\cos B$ is equal to

• A. $\frac{17}{35}$
• B. $\frac{17}{70}$
• C. $\frac{19}{35}$
• D. $\frac{19}{70}$

Hint:

In ratio problems, use a constant $k$ to represent side lengths and then apply standard triangle formulae.

Answer 1

The Correct Option is C

Step 1: Find ratios of sides
Let $\frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13} = k$.
$b+c = 11k, c+a = 12k, a+b = 13k$.
Summing: $2(a+b+c) = 36k \implies a+b+c = 18k$.

Step 2: Calculate individual sides
$a = (a+b+c) - (b+c) = 18k - 11k = 7k$.
$b = (a+b+c) - (c+a) = 18k - 12k = 6k$.
$c = (a+b+c) - (a+b) = 18k - 13k = 5k$.

Step 3: Use Cosine Rule
$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{(7k)^2 + (5k)^2 - (6k)^2}{2(7k)(5k)}$.
$\cos B = \frac{49 + 25 - 36}{70} = \frac{38}{70} = \frac{19}{35}$.
Final Answer: (C)