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  4. Given (b+c/11) = (c+a/12)= (a+b/13) for a Δ ABC with usual notation. If ( cos A/α) = ( cos B/β) = ( cos C/γ) , then the ordered triad ( α, β, γ) has a value :

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Q. 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 :

JEE MainJEE Main 2019Trigonometric Functions

Solution:

$b + c = 11 \lambda , c + a = 12 \lambda, a + b = 13 \lambda$
$\Rightarrow \; a = 7 \lambda , b = 6 \lambda , c = 5\lambda$
(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 $