Question

In the given figure, ABC is a triangle , in which $\angle BAC$ = 30 . Show that the length of BC is equal to the radius of the circumcircle of ABC , whose centre is O .

Answer 1

The angle subtended by a chord at center is double the angle subtended by it on any other point on the major segment of circle.

So $\angle BOC={60}^{\circ }\Delta BOCOB=OC$

The sides are equal and included angle is ${60}^{\circ }$

So the triangle is eqquilateral

Hence $BC=OB$(radius)