MHT CET · Maths · Properties of Triangles
In a triangle \(A B C\), with usual notations, if \(\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}\), then \(\cos A: \cos B: \cos C=\)
- A 11:12:13
- B 25:19:7
- C 7:19:25
- D \(19: 7: 25\)
Answer & Solution
Correct Answer
(C) 7:19:25
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=k(\text { say }) \\ & \Rightarrow a+b+c=18 k \\ & \Rightarrow a=7 k, b=6 k, c=5 k\end{aligned}\)
Using cosine rule
\(\begin{aligned} & \cos A=\frac{1}{5}, \cos B=\frac{19}{35}, \cos C=\frac{5}{7} \\ & \Rightarrow \cos A: \cos B: \cos C=\frac{1}{5}: \frac{19}{35}: \frac{5}{7}=7: 19: 25\end{aligned}\)
Using cosine rule
\(\begin{aligned} & \cos A=\frac{1}{5}, \cos B=\frac{19}{35}, \cos C=\frac{5}{7} \\ & \Rightarrow \cos A: \cos B: \cos C=\frac{1}{5}: \frac{19}{35}: \frac{5}{7}=7: 19: 25\end{aligned}\)
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