Question

An equilateral triangle ABC is inscribed in a circle with centre O. Then, $\angle BOC$ is equal to ___.

• A. 30${}^{\circ }$
• B. 60${}^{\circ }$
• C. 90${}^{\circ }$
• D. 120${}^{\circ }$

Answer 1

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

Given that $\Delta ABC$ is equilateral.
$⟹\angle BAC={60}^{\circ }$
Since the angle subtended by a chord at the centre of a circle is twice the angle subtended by the same chord at any other point on the remaining part of the circle, we have
$\angle BOC=2\angle BAC=2\times {60}^{\circ }={120}^{\circ }.$