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MHT CET202519 Apr 2025Morning ShiftMathematicsProperties of TrianglesActual

In a triangle A B C , with usual notations, if b+c 11 = c+a 12 = a+b 13 Then A: B: C is

Options

  1. A7: 19: 25
  2. B19: 7: 25
  3. C12: 14: 20
  4. D19: 25: 20

Correct answer

A. 7: 19: 25

Step-by-step solution

Let the given ratio equal k , so b+c 11 = c+a 12 = a+b 13 = k . This gives the system of equations: b+c = 11k c+a = 12k a+b = 13k Adding these equations yields 2(a+b+c) = 36k , so a+b+c = 18k . Subtracting each original equation from this sum gives the side lengths: a = (a+b+c) - (b+c) = 18k - 11k = 7k b = 18k - 12k = 6k c = 18k - 13k = 5k Using the cosine rule A = b^2+c^2-a^2 2bc , and similarly for B and C , with substitution: A = (6k)^2+(5k)^2-(7k)^2 2(6k)(5k) = 36+25-49 60 = 12 60 = 1 5 B = (7k)^2+(5k)^2-(6k)^2

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