MHT CET · Maths · Properties of Triangles
With usual notation, in a triangle ABC \(\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}\), then the value of \(\cos B\) is equal to
- A \(\frac{17}{35}\)
- B \(\frac{17}{70}\)
- C \(\frac{19}{35}\)
- D \(\frac{19}{70}\)
Answer & Solution
Correct Answer
(C) \(\frac{19}{35}\)
Step-by-step Solution
Detailed explanation
\(b+c=11k, c+a=12k, a+b=13k\) \(2(a+b+c) = (11+12+13)k = 36k \Rightarrow a+b+c = 18k\)
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