Question

In the figure, ABC is a triangle in which $\angle BAC={30}^{\circ }$. Show that BC is equal to the radius of the circumcircle of $\Delta ABC$, whose centre is O.
[3 MARKS]

Answer 1

Concept: 1 Mark
Steps: 1 Mark
Answer: 1 Mark

Join OB and OC.
The angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.

$\angle OBC+\angle OCB+\angle BOC={180}^{\circ }$

$2\angle OBC+{60}^{\circ }={180}^{\circ }$

$2\angle OBC={120}^{\circ }$

$\angle OBC={60}^{\circ }$

Thus, we have

$\angle OBC=\angle OCB=\angle BOC={60}^{\circ }$

$\Delta OBC$ is an equilateral triangle.

$\therefore$ BC is equal to the radius of the circumcircle of $\Delta ABC$.