A) Equilateral
B) Isosceles
C) Right angled
D) None of these
Correct Answer: C
Solution :
We have, \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ac-ab\sqrt{3}=0\] \[\frac{{{a}^{2}}}{4}-\,ac+{{c}^{2}}+\frac{3{{a}^{2}}}{4}+{{b}^{2}}-ab\sqrt{3}=0\] \[{{\left[ \frac{a}{2}-c \right]}^{2}}+{{\left[ \frac{\sqrt{3}a}{2}-b \right]}^{2}}=0\] i.e, \[a=2c\] and \[2b=\sqrt{3}a\] i.e., \[{{b}^{2}}+{{c}^{2}}={{a}^{2}}\] Hence triangle is right angled.
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