Question

6. If cos A + cos B + cos C = 3/2, then show that the triangle is equilateral​

Answer 1

Answer:

Let a, b, c are the sides of a ∆ ABC. Given, cos A + cos B + cos C = 3/2

⇒ (b2 + c2 - a2)/2bc + (a2 + c2 - b2)/2ac + (a2 + b2 - c2)/2ab = 3/2

⇒ ab2 + ac2 – a3 + ba2 + bc2 – b3 + ca2 + cb2 – c3 = 3abc

⇒ a(b – c)2 + b(c – a)2 + c(a – b)2 = (a + b + c)/2[(a - b)2 + (b - c)2 + (c - a)2]

⇒ (a + b – c) (a – b)2 + (b + c – a) (b – c)2 + (c + a – b)

=> (c – a)^2 = 0

(as we know, a+b–c > 0, b+c–a > 0, c+a–b > 0)

∴ Each term on the left of equation has positive coefficient multiplied by perfect square, each term must be separately zero. ⇒ a = b = c

∴ Triangle is an equilateral