Properties of Triangles 1 Question 2
2. Given, $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$ for a $\triangle A B C$ with usual notation. If $\frac{\cos A}{\alpha}=\frac{\cos B}{\beta}=\frac{\cos C}{\gamma}$, then the ordered $\operatorname{triad}(\alpha, \beta, \gamma)$ has a value
(2019 Main, 11 Jan II)
(a) $(19,7,25)$
(b) $(3,4,5)$
(c) $(5,12,13)$
(d) $(7,19,25)$
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Answer:
Correct Answer: 2. (d)
Solution:
- Given, $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=\lambda$ (say)
$$ \begin{aligned} & b+c=11 \lambda, c+a=12 \lambda \text { and } a+b=13 \lambda \\ & \Rightarrow \quad 2(a+b+c)=36 \lambda \\ & \Rightarrow \quad a+b+c=18 \lambda \end{aligned} $$
From Eqs. (i) and (ii), we get
$$ a=7 \lambda, b=6 \lambda, c=5 \lambda $$
Now,
$$ \begin{aligned} & \cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{\lambda^{2}[36+25-49]}{60 \lambda^{2}}=\frac{12}{60}=\frac{1}{5} \\ & \begin{aligned} \cos B= & \frac{a^{2}+c^{2}-b^{2}}{2 a c}=\frac{\lambda^{2}[49+25-36]}{70 \lambda^{2}}=\frac{19}{35} \\ \text { and } \cos C & =\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{\lambda^{2}[49+36-25]}{84 \lambda^{2}} \\ & =\frac{60}{84}=\frac{5}{7} \end{aligned} \\ & \text { Thus, } \cos A=\frac{1}{5}=\frac{7}{35}, \\ & \qquad \cos B=\frac{19}{35}, \cos C=\frac{25}{35} \\ & \frac{\cos A}{7}=\frac{\cos B}{19}=\frac{\cos C}{25}=\frac{1}{35} \\ & \Rightarrow \quad(\alpha, \beta, \gamma)=(7,19,25) \end{aligned} $$
