Given, for a with usual notation. If , then the ordered triad has a value
- A.(19, 7, 25)
- B.(3, 4, 5)
- C.(5, 12, 13)
- D.(7, 19, 25)
Show answer and solution
Correct option: D
Solution
Let k = (b+c)/11 = (c+a)/12 = (a+b)/13.
Adding all three: 2(a+b+c) = 36k, so a+b+c = 18k.
Then: a = 18k - 11k = 7k, b = 18k - 12k = 6k, c = 18k - 13k = 5k.
So sides are in ratio a:b:c = 7:6:5.
Using cosine rule:
cosA = (b²+c²-a²)/(2bc) = (36+25-49)/(2·6·5) = 12/60 = 1/5
cosB = (a²+c²-b²)/(2ac) = (49+25-36)/(2·7·5) = 38/70 = 19/35
cosC = (a²+b²-c²)/(2ab) = (49+36-25)/(2·7·6) = 60/84 = 5/7
So cosA : cosB : cosC = (1/5) : (19/35) : (5/7) = 7 : 19 : 25 (multiply through by 35).
Therefore (α, β, γ) = (7, 19, 25).
Adding all three: 2(a+b+c) = 36k, so a+b+c = 18k.
Then: a = 18k - 11k = 7k, b = 18k - 12k = 6k, c = 18k - 13k = 5k.
So sides are in ratio a:b:c = 7:6:5.
Using cosine rule:
cosA = (b²+c²-a²)/(2bc) = (36+25-49)/(2·6·5) = 12/60 = 1/5
cosB = (a²+c²-b²)/(2ac) = (49+25-36)/(2·7·5) = 38/70 = 19/35
cosC = (a²+b²-c²)/(2ab) = (49+36-25)/(2·7·6) = 60/84 = 5/7
So cosA : cosB : cosC = (1/5) : (19/35) : (5/7) = 7 : 19 : 25 (multiply through by 35).
Therefore (α, β, γ) = (7, 19, 25).
[JEE Main 2019 · 11 Jan · Shift II]