Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

If $a:b:c$ $=\sqrt{3}:1:1,$ then the triangle is?

• A. A right angled triangle
• B. An obtuse angled triangle
• C. An acute angled triangle, that is not isosceles
• D. An equilateral triangle

Answer 1

Answer: (B)

An obtuse angled triangle

Given, in a triangle ABC $a:b:c=\sqrt{3}:1:1$
the angle opposite to largest side is the largest angle.
$\cos A=\frac{b^2+c^2-a^2}{2bc}$

$\cos A=\frac{1+1-3}{2}$
$\angle A={120}^{\circ }$

$\therefore$ $\Delta ABC$ is an obtuse angled triangle.