Question

The lengths of the sides of a triangle are $(x-y),(x+y)$ and $\sqrt{3x^2+y^2},x,y>0$. Then the largest angle of the triangle is $\frac{2\pi }{3}$

• A. True
• B. False

Answer 1

The correct option is A True

Evidently $\sqrt{3x^2+y^2}$ is the largest side and therefore $\angle C$ is the greatest angle.
$\cos C=\frac{a^2+b^2-c^2}{2ab}$
$=\frac{{(x-y)}^2+{(x+y)}^2-{\sqrt{(3x^2+y^2)}}^2}{2(x-y)(x+y)}$
$=\frac{x^2+y^2-2xy+x^2+y^2+2xy-3x^2-y^2}{2(x^2-y^2)}$
$=\frac{-x^2+y^2}{2(x^2-y^2)}$
$=\frac{-(x^2-y^2)}{2(x^2-y^2)}=\frac{-1}{2}$
$\therefore \angle C=\frac{2\pi }{3}$