# With usual national, if in a triangle ABC;b + c/11 = c + a/12 = a + b/13 then prove that cos A/7 = cos B/19 = cos C/25.

7 years ago
Hello Student,
Given that, in ∆ABC,
b + c/11 = c + a/12 = a + b/13
where a, b, c are the lengths of sides, BC, CA and AB respectively.
Let b + c/11 = c + a/12 = a + b/13 = k
⇒ b + c = 11 k . . . . . . . . . . . . . . . . . . . . . (1)
⇒ c + a = 11 k . . . . . . . . . . . . . . . . . . . . . (2)
⇒ a + b = 13 k . . . . . . . . . . . . . . . . . . . . . (3)
Adding the above three eqs. We get
2 (a + b + c) = 36 k
⇒ a + b + c = 18 k . . . . . . . . . . . . . . . . . . . . (4)
Solving each of (1), (2) and (3) with (4), we get
Now, cos A = b2 + c2 – a2/2bc
= 36 k2 + 25 k2 – 49 k2/2 x 6k x 5k = 12 k2/60 k2 = 1/5
Cos B = c2 + a2 – b2/2ca = 25 k2 + 49 – 36 k2/2 x 5 k x 7 k
= 38 k2/70 k2 = 19/35
Cos C = a2 + b2 – c2/2ab = 49 k2 + 36k2 – 25k2/2 x 7 k x 6 k
= 60 k2/84 k2 = 5/7
∴ cos A / 1/5 = cos B / 19/35 = cos C / 5/7
⇒ cos A / 7/35 = cos B / 19/35 = cos C / 25/35
⇒ cos A/7 = cos B/19 = cos C/25

Thanks