Properties of Triangles 1 Question 22
22. With usual notation, if in a $\triangle A B C \frac{b+c}{11}=\frac{c+a}{12}$ $=\frac{a+b}{13}$, then prove that $\frac{\cos A}{7}=\frac{\cos B}{19}=\frac{\cos C}{25}$.
(1984, 4M)
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Solution:
- Let $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=\lambda$
$\Rightarrow(b+c)=11 \lambda, c+a=12 \lambda, a+b=13 \lambda$
$\Rightarrow \quad 2(a+b+c)=36 \lambda$
$$ \Rightarrow \quad a+b+c=18 \lambda $$
On solving Eqs. (i) and (ii), we get
$$ \begin{gathered} a=7 \lambda, b=6 \lambda \text { and } c=5 \lambda \\ \cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{36 \lambda^{2}+25 \lambda^{2}-49 \lambda^{2}}{2(30) \lambda^{2}}=\frac{1}{5} \\ \cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}=\frac{49 \lambda^{2}+25 \lambda^{2}-36 \lambda^{2}}{70 \lambda^{2}}=\frac{19}{35} \\ \cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{49 \lambda^{2}+36 \lambda^{2}-25 \lambda^{2}}{84 \lambda^{2}}=\frac{5}{7} \end{gathered} $$
$\therefore \cos A: \cos B: \cos C=\frac{1}{5}: \frac{19}{35}: \frac{5}{7}=7: 19: 25$
