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)
Show Answer
Solution:
- 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.
