Question

An equilateral $△ABC$ is inscribed in a circle with centre $O$. Find the measure of $\angle BOC$.

Answer 1

We have,

$\Delta ABC$ is an equilateral triangle

Since, each angle is ${60}^o$.

Now, using property Angle subtend at the centre of circle is double angle at any point on circumference of circle

$\angle BOC=2\angle BAC$

$\angle BOC=2\times {60}^o$

$\angle BOC={120}^o$

Hence, this is the answer.