Question:
Given $\frac{\mathrm{b}+\mathrm{c}}{11}=\frac{\mathrm{c}+\mathrm{a}}{12}=\frac{\mathrm{a}+\mathrm{b}}{13}$ for a $\triangle \mathrm{ABC}$ with
usual notation. If $\frac{\cos \mathrm{A}}{\alpha}=\frac{\cos \mathrm{B}}{\beta}=\frac{\cos \mathrm{C}}{\gamma}$, then
the ordered triad $(\alpha, \beta, \gamma)$ has a value :-
Correct Option: , 3
Solution:
$\mathrm{b}+\mathrm{c}=11 \lambda, \mathrm{c}+\mathrm{a}=12 \lambda, \mathrm{a}+\mathrm{b}=13 \lambda$
$\Rightarrow a=7 \lambda, b=6 \lambda, c=5 \lambda$
(using cosine formula)
$\cos \mathrm{A}=\frac{1}{5}, \cos \mathrm{B}=\frac{19}{35}, \cos \mathrm{C}=\frac{5}{7}$
$\alpha: \beta: \gamma \Rightarrow 7: 19: 25$