Question

If in $△ABC,\cos A+\cos B+\cos C=\frac{3}{2}$, then triangle $△ABC$ is?

• A. Right angled
• B. Isosceles
• C. Acute
• D. Equilateral

Answer 1

The correct option is D Equilateral

$\because \cos A+\cos B+\cos C=\frac{3}{2}$
$\Rightarrow \frac{b^2+c^2-a^2}{2bc}+\frac{a^2+c^2-b^2}{2ac}+\frac{a^2+b^2-c^2}{2ab}=\frac{3}{2}$
$\Rightarrow (a+b-c)(a-b{)}^2+(b+c-a\left)\right(b-c{)}^2+(c+a-b)(c-a{)}^2=0$
As we know, $a+b>c,b+c>a,c+a>b$
$\Rightarrow a=b=c$
Hence, triangle is an equilateral.