Question

lf in a triangle $ABC,a^2+b^2+c^2=ca+ab\sqrt{3}$ , then the triangle is

• A. equilateral
• B. right angled and isosceles
• C. right angled with one of the acute angles measuring $\frac{\pi }{3}$
• D. obtuse angled

Answer 1

The correct option is C right angled with one of the acute angles measuring $\frac{\pi }{3}$

Given, $a^2+b^2+c^2=ac+ab\sqrt{3}$

$=a(c+b\sqrt{3})$

$=2a(\frac{1}{2}c+b\frac{\sqrt{3}}{2})$

$=2a(c\cos {60}^0+b\cos {30}^0)=2a(c\cos B+b\cos C)$

Hence

$\angle B={60}^0$ and $\angle C={30}^0$

Therefore

$\angle A={90}^0$.