Question

In a $\Delta ABC$, if $a^2+b^2+c^2=ac+\sqrt{3}ab$, then the triangle is

• A. equilateral
• B. right angled and isosceles
• C. right angled and not isosceles
• D. None of the above

Answer 1

The correct option is C right angled and not isosceles

$a^2=b^2+c^2-2bc\cos A$
$b^2=a^2+c^2-2ac\cos B$
$c^2=b^2+a^2-2ba\cos C$
Adding above three equation and rearranging
$2bc\cos A+2ac\cos B+2ab\cos C=a^2+b^2+c^2$
From the question
$ac+\sqrt{3}ab=a^2+b^2+c^2$
$⟹\cos A=0,\cos B=\frac{1}{2},\cos C=\frac{\sqrt{3}}{2}$
So $\angle A=90$ degrees
So $△ABC$ is right angles and not isoceles