Question

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

• A. $\frac{9}{10}$
• B. $\frac{10}{11}$
• C. $\frac{11}{12}$
• D. $\frac{12}{13}$

Answer 1

Answer: (C)

$\frac{11}{12}$

Answer 2

Let $\frac{b +c}{9}=\frac{c +a}{10}=\frac{a +b}{11}=k$
$\Rightarrow b+ c=9 k, c +a=10 k$ and $a+b=11\, k$
and $a+b+c=15\, k$
$\therefore a=6 k,\, b=5\, k$ and $c=4 k$
$\because \frac{\cos A+\cos B}{\cos C}=\frac{\frac{b^{2}+c^{2}-a^{2}}{2 b c}+\frac{a^{2}+c^{2}-b^{2}}{2 a c}}{\frac{a^{2}+b^{2}-c^{2}}{2 a b}}$
$=\frac{\frac{25+16-36}{40}+\frac{36+16-25}{48}}{\frac{36+25-16}{60}}$
$=\frac{\frac{5}{40}+\frac{27}{48}}{\frac{45}{60}}$
$=\frac{\frac{1}{8}+\frac{9}{16}}{\frac{3}{4}} \frac{=\frac{11}{16}}{\frac{3}{4}}=\frac{11}{12}$