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MHT CET20255 May 2025Evening ShiftMathematicsProperties of TrianglesActual

With usual notation, in a triangle ABC b+c 11 = c+a 12 = a+b 13 , then the value of B is equal to

Options

  1. A17 35
  2. B17 70
  3. C19 35
  4. D19 70

Correct answer

C. 19 35

Step-by-step solution

Let the common ratio be k , giving the system: b + c 11 = c + a 12 = a + b 13 = k b + c = 11k c + a = 12k a + b = 13k Adding yields: (b + c) + (c + a) + (a + b) = 11k + 12k + 13k 2(a + b + c) = 36k a + b + c = 18k Solving for each variable: a = (a + b + c) - (b + c) = 18k - 11k = 7k b = (a + b + c) - (a + c) = 18k - 12k = 6k c = (a + b + c) - (a + b) = 18k - 13k = 5k Applying the Law of Cosines : B = a^2 + c^2 - b^2 2ac B = (7k)^2 + (5k)^2 - (6k)^2 2(7k)(5k) = 49k^2 + 25k^2 - 36k^2 70k^2 = 38k^2 70k^2 = 19 35 The r

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