Question

In the given figure,

If $O$ is the center of the circle and $\angle AOB=\angle BOC={120}^o$, then $△ABC$ is

• A. Equilateral triangle
• B. Right angled triangle
• C. Isosceles triangle
• D. Scalene triangle

Answer 1

Answer: (A), (C)

Isosceles triangle


In the above figure, $O$ is center of circle and $\angle AOB=\angle BOC={120}^o$
We know that chords subtending equal angles at the center are also equal, so $AB=BC$

Also,
$\angle AOB+\angle BOC+\angle AOC={360}^o{120}^o+{120}^o+\angle AOC={360}^o{240}^o+\angle AOC={360}^o\angle AOC={360}^o-{240}^o\angle AOC={120}^o$

As, $\angle BOC=\angle AOC={120}^o$
So, $BC=AC$

So, $AB=BC=AC$
$△ABC$ is an equilateral triangle.
And all equilateral triangles are Isosceles triangle so $△ABC$ is an isosceles triangle also.

Hence, option $\left(a\right)$ and $\left(c\right)$ correct.