Question

If the sides of a triangle ABC, are a, b, $\sqrt{a^2+b^2+ab}$ then the greatest angle of the triangle is -

• A. ${160}^{\circ }$
• B. ${150}^{\circ }$
• C. ${120}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is C ${120}^{\circ }$

Here, we are given all three sides and we have to find something about angles so we’ll use cosines formula.
$Cos\left(\theta \right)=\frac{a^2+b^2-(\sqrt{a^2+b^2+ab}{)}^2}{2.a.b}$ where $\theta$ is the angle opposite to side length of
$\sqrt{a^2+b^2+ab}$
$Cos\left(\theta \right)=\frac{-a.b}{2.a.b}$
$Cos\left(\theta \right)=\frac{-1}{2}$
$\theta ={120}^{\circ }$
Since, one angle of a triangle is ${120}^{\circ }$, the sum of other two angles would be ${60}^{\circ }$ making ${120}^{\circ }$ as the greatest angle.