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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 :-

  1. $(3,4,5)$

  2. $(19,7,25)$

  3. $(7,19,25)$

  4. $(5,12,13)$


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)

$\cos \mathrm{A}=\frac{1}{5}, \cos \mathrm{B}=\frac{19}{35}, \cos \mathrm{C}=\frac{5}{7}$

$\alpha: \beta: \gamma \Rightarrow 7: 19: 25$

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