In triangle ABC,if frac{b+c}{9}=frac{c+a}{10} | vedclass

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(C) Let $\frac{b+c}{9}=\frac{c+a}{10}=\frac{a+b}{11}=k$. Then $b+c=9k$,$c+a=10k$,and $a+b=11k$. Adding these gives $2(a+b+c)=30k$,so $a+b+c=15k$. Thus

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$\frac{11}{12}$

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(C) Let $\frac{b+c}{9}=\frac{c+a}{10}=\frac{a+b}{11}=k$. Then $b+c=9k$,$c+a=10k$,and $a+b=11k$. Adding these gives $2(a+b+c)=30k$,so $a+b+c=15k$. Thus,$a=(a+b+c)-(b+c)=15k-9k=6k$,$b=(a+b+c)-(c+a)=15k-10k=5k$,and $c=(a+b+c)-(a+b)=15k-11k=4k$. Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{25k^2+16k^2-36k^2}{2(5k)(4k)} = \frac{5k^2}{40k^2} = \frac{1}{8}$. $\cos B = \frac{a^2+c^2-b^2}{2ac} = \frac{36k^2+16k^2-25k^2}{2(6k)(4k)} = \frac{27k^2}{48k^2} = \frac{9}{16}$. $\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{36k^2+25k^2-16k^2}{2(6k)(5k)} = \frac{45k^2}{60k^2} = \frac{3}{4}$. Therefore,$\frac{\cos A+\cos B}{\cos C} = \frac{\frac{1}{8}+\frac{9}{16}}{\frac{3}{4}} = \frac{\frac{2+9}{16}}{\frac{3}{4}} = \frac{11}{16} 	imes \frac{4}{3} = \frac{11}{12}$. Hence,option $C$ is correct.

In triangle $ABC$,if $\frac{b+c}{9}=\frac{c+a}{10}=\frac{a+b}{11}$,then $\frac{\cos A+\cos B}{\cos C}=$

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