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For a triangle ABC it is given that cosA + cosB + cosC = 3/2.then triangle is

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4 months ago

```							Let a, b, c are the sides of a ∆ ABC. Given, cos A + cos B + cos C = 3/2 ⇒ (b^2 + c^2 - a^2)/2bc + (a^2 + c^2 - b^2)/2ac + (a^2 + b^2 - c^2)/2ab = 3/2 ⇒ ab^2 + ac^2 – a^3 + ba^2 + bc^2 – b^3 + ca^2 + cb^2 – c^3 = 3abc ⇒ a(b – c)^2 + b(c – a)^2 + c(a – b)^2 = (a + b + c)[(a - b)^2 + (b - c)^2 + (c - a)^2]/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 equilateral.KINDLY APPROVE:))
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4 months ago
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