Question

The length of the sides of a triangle are $x,y$ and $\sqrt{x^2+y^2+xy}$ and The measure of the greatest angle is

• A. 120º
• B. 150º
• C. 135º
• D. None of these

Answer 1

The correct option is A

120º

Let $a=x,b=yandc=\sqrt{x^2+y^2+xy}$
Then $a^2=x^2,b^2=y^2,and{c}^2=x^2+y^2+xy$
$\therefore$ C is the Length of the greatest side and Hence $\angle C$ is the greatest angle
$\therefore$ By Cosine Rule
$cosC=\frac{a^2+b^2-c^2}{2ab}=\frac{x^2+y^2-x^2-y^2-xy}{2xy}$
$=-\frac{1}{2}$ $\therefore \angle C={120}^o$