Question

If the sides of a triangle are in the ratio $x:y:\sqrt{x^2+xy+y^2}$, then the greatest angle is:

• A. ${90}^o$
• B. ${120}^o$
• C. ${\cos }^{-1}\left(\frac{x+y}{x-y}\right)$
• D. ${30}^o$

Answer 1

The correct option is B ${120}^o$

Given Ratio of sides is $x:y:\sqrt{x^2+y^2+xy}$

Let $a=x,b=y,c=\sqrt{x^2+xy+y^2}$

$c$ is the largest side

$\cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{x^2+y^2-(x^2+y^2+xy)}{2xy}=-\frac{1}{2}⟹C={120}^{\circ }$