With usual notation, if in triangle ABC, b+c11=c+a12=a+b13, then cosA:cosB:cosC=

Let b+c11=c+a12=a+b13=k; a+b+c=18k; a=7k; b=6k; c=5kcosA:cosB:cosC=7:19:25

With usual notation, if in triangle ABC, b+c11=c+a12=a+b13, then cosA:cosB:cosC=

Let b+c11=c+a12=a+b13=k; a+b+c=18k; a=7k; b=6k; c=5kcosA:cosB:cosC=7:19:25

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With usual notation, if in triangle ABC, $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13}$ , then $\cos A : \cos B : \cos C =$

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7 : 19 : 25

19 : 25 : 7

19 : 7 : 25

7 : 25 : 19

answer is A.

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Detailed Solution

Let $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13} = k; a + b + c = 18 k$ ; $a = 7 k; b = 6 k; c = 5 k$

$\cos A : \cos B : \cos C = 7 : 19 : 25$

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With usual notation, if in triangle ABC, b+c11=c+a12=a+b13, then cosA:cosB:cosC=

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With usual notation, if in triangle ABC, $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13}$ , then $\cos A : \cos B : \cos C =$