Question

lf $CotA+CotB+CotC=\sqrt{3}$, then the triangle $ABC$ is

• A. isosceles
• B. equilateral
• C. right angled
• D. scalene

Answer 1

The correct option is B equilateral

$cot\Delta =\frac{a^2+c^2-b^2}{4\Delta }$
$=\frac{a^2+b^2-c^2}{4\Delta }+\frac{b^2+c^2-{\Delta }^2}{4\Delta }+\frac{a^2+c^2-b^2}{4\Delta }$
$\frac{a^2+b^2+c^2}{4\Delta }=\sqrt{3}$
$a=b=c$
$=\frac{3\times a^2}{4\frac{a^2\sqrt{3}}{4}}$
$=\sqrt{3}$