Question

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

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

Answer 1

Answer: (C)

(7, 19, 25)

Answer 2

$b + c = 11 \lambda , c + a = 12 \lambda, a + b = 13 \lambda$
$\Rightarrow \; a = 7 \lambda , b = 6 \lambda , c = 5\lambda$
(using cosine formula)
$\cos A =\frac{1}{5} , \cos B = \frac{19}{35} , \cos C = \frac{5}{7}$
$\alpha: \beta:\gamma \Rightarrow 7 : 19 : 25$