Question

In a triangle, $a^2+b^2+c^2=ca+ab\sqrt{3}$. Then the triangle is:

• A. Equilateral
• B. Right angled and isosceles
• C. Right angled with $\angle A={90}^o,\angle B={60}^o$ and $\angle C={30}^o$
• D. None of these

Answer 1

The correct option is C Right angled with $\angle A={90}^o,\angle B={60}^o$ and $\angle C={30}^o$

Given: $a^2+b^2+c^2=ca+ab\sqrt{3}$

$\Rightarrow b^2-b\left(a\sqrt{3}\right)+(a^2+c^2-ac)=0$

Descriminant $\Delta =3a^2-4(a^2+c^2-ac)$

$\Rightarrow \Delta =-(a-2c{)}^2$

$\Rightarrow \Delta \le 0$

But we know that the side $b$ is real

$\Rightarrow \Delta =0$

$\Rightarrow a=2c$

Substituting in above equation ;

$\Rightarrow b^2+5c^2=2c^2+2bc\sqrt{3}$

$\Rightarrow 3c^2-2bc\sqrt{3}+b^2=0$

$\Rightarrow b=c\sqrt{3}$

$\therefore a^2=b^2+c^2$

$\therefore$ it is a right angled

Also, $\sin \angle A={90}^o,\sin \angle B=\frac{b}{a}=\frac{c\sqrt{3}}{2c}=\frac{\sqrt{3}}{2}\Rightarrow \angle B={60}^o$ and

$\sin \angle C=\frac{c}{a}=\frac{c}{2c}=\frac{1}{2}\Rightarrow \angle C={30}^o$

Hence, the triangle $ABC$ is a right angled triangle with $\angle A={90}^o,\angle B={60}^o$ and $\angle C={30}^o$