Properties of Triangles 2 Question 14

14. For a $\triangle A B C$, it is given that $\cos A+\cos B+\cos C=\frac{3}{2}$. Prove that the triangle is equilateral.

(1984, 4M)

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Solution:

  1. Let $a, b, c$ are the sides of a $\triangle A B C$.

Given, $\quad \cos A+\cos B+\cos C=\frac{3}{2}$

$\Rightarrow \frac{b^{2}+c^{2}-a^{2}}{2 b c}+\frac{a^{2}+c^{2}-b^{2}}{2 a c}+\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{3}{2}$

$\Rightarrow \quad a b^{2}+a c^{2}-a^{3}+b a^{2}+b c^{2}-b^{3}$

$$ +c a^{2}+c b^{2}-c^{3}=3 a b c $$

$\Rightarrow a(b-c)^{2}+b(c-a)^{2}+c(a-b)^{2}$

$$ =\frac{(a+b+c)}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] $$

$\Rightarrow(a+b-c)(a-b)^{2}+(b+c-a)(b-c)^{2}$

$$ +(c+a-b)(c-a)^{2}=0 $$

[as we know, $a+b-c>0, b+c-a>0, c+a-b>0$ ] $\therefore$ Each term on the left of equation has positive coefficient multiplied by perfect square, each term must be separately zero.

$\Rightarrow$

$$ a=b=c $$

$\therefore$ Triangle is an equilateral.



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