In triangle $A B C$, 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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Aicharya · Accepted Answer

$\begin{aligned} & \text {Let } \frac{b+c}{9}=\frac{c+a}{10}=\frac{a+b}{11}=k \ & \Rightarrow b+c=9 k, c+a=10 k \text { and } a+b=11 k \ & \text { and } a+b+c=15 k \ & \therefore a=6 k, b=5 k \text { and } c=4 k \ & \because \frac{\cos A+\cos B}{\cos C}=\frac{\frac{b^2+c^2-a^2}{2 b c}+\frac{a^2+c^2-b^2}{2 a c}}{\frac{a^2+b^2-c^2}{2 a b}} \ & =\frac{\frac{25+16-36}{40}+\frac{36+16-25}{48}}{\frac{36+25-16}{60}} \ & =\frac{\frac{5}{40}+\frac{27}{48}}{\frac{45}{60}} \ & =\frac{\frac{1}{8}+\frac{9}{16}}{\frac{3}{4}} \frac{\frac{11}{16}}{\frac{3}{4}}=\frac{11}{12} \end{aligned}$ Hence, option (3) is correct.

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

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