Question

REQ. URGENTLY PLS SOLVE . WRONG ANSWERS WILL BE REPORTED ️️​

Answer 1

Step-by-step explanation:

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.

HENCE PROVE.