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# For a triangle ABC it is given that cos A + cos B + cos C = 3/2 Prove that the triangle is equilateral.

7 years ago
Hello Student,
Given that in ∆ ABC
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 + a2b + bc2 – b3 + ac2 + b2 c – c3 = 3 abc
⇒ ab2 + ac2 + bc2 + ba2 + ca2 + cb2 - 6abc = a3 + b3 + 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 . . . . . . . . . . . . . . . . . . . (1)
a + b > c
b +c > a {sum of any two sides of a ∆ is greater than the third side
As we know that c + a > b
∴ Each part on the LHS of eq. (1) has +ve coeff. Multiplied by perfect square, each must be separately zero
∴ a – b = 0 ; b – c = 0 ; c – a = 0 ⇒ a = b = c
Hence ∆ is an equilateral ∆
ALTERNATE SOLUTION :
Given that cos A + cos B + cos C = 3/2 in ∆ABC
⇒ 2 cos A + B/2 cos A – B/2 = 3/2 – cos C
⇒ 2 sin C/2 cos A – B/2 = 3 – 2 cos C/2
⇒ 2 sin C/2 cos A – B/2 = 3 – 2(1 – 2 sin2 C/2/4 sin C/2
⇒ cos (A – B/2) = 1 + 4 sin2 C/2 / 4 sin C/2
⇒ cos (A – B/2) = 1 + 4 sin2 C/2 – 4 sin C/2 + 4 sin C/2 / 4 sin C/2
⇒ cos( A – B/2) = (1 – 2 sin C/2)2/4 sin C/2 + 1
Which is possible only when
1 – 2 sin C/2 = 0 ⇒ sin C/2 = 1/2
⇒ C/2 = 30° ⇒ C = 60°
Also then cos A – B/2 = 1 ⇒ A – B/2 = 0 ⇒ A – B=0 . . . . . . . . . . . . . . . . . . . . . (1)
And A + B = 180° - 60° = 120° A + B = 120° . . . . . . . . . . . . . . . . . . . . . . . . (2)
From (1) and (2) A = B = 60°
Thus we get A = B = C = 60° ∴ ∆ ABC is an equilateral ∆.

Thanks