Question

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

• A. equilateral
• B. isosceles
• C. right angled
• D. none of these

Answer 1

Answer: (C)

right angled

given that, $a^2+b^2+c^2=ac+ab\sqrt{3}$
$\Rightarrow {(\frac{\sqrt{3}a}{2}-b)}^2+{(\frac{a}{2}-c)}^2=0$
Therefore, $\frac{\sqrt{3}a}{2}=b$ and $\frac{a}{2}=c$
$\Rightarrow a=\frac{b}{\sqrt{3}/2}=\frac{c}{1/2}$
$\Rightarrow \frac{a}{\sin \frac{\pi }{2}}=\frac{b}{\sin \frac{\pi }{3}}=\frac{c}{\sin \frac{\pi }{6}}$
Therefore, the given triangle is an right angled.