Question

An equilateral triangle ABC is inscribed in a circle with centre O. Find $\angle BOC$.

Answer 1

Given that $\Delta ABC$ is an equilateral triangle.
$\therefore \angle BAC={60}^{\circ }$
Using the property, angle subtended by the chord at the centre is twice the angle subtended by it on the circle,
we get,
$\Rightarrow \angle BOC=2\angle BAC$
$\Rightarrow \angle BOC=2\times {60}^{\circ }$
$\therefore \angle BOC={120}^{\circ }$