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Angle Subtended by an Arc of a Circle on the Circle and at the Center

In the given ...

Question

In the given figure, ABC is a triangle in which $∠ B A C = 30^{\circ} .$ Show that BC is equal to the radius of the circumcircle of the triangle ABC, whose centre is O.

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Solution

Given $\angle BAC = 30^o$

$\angle BOC = 2 \angle BAC = 2 . 30 = 60^o$ (angle at the centre of the circle is twice that of the angle subtended at the circumference by the same chord)

In $\triangle OBC$

$OB = OC$ (radius of the same circle)

$\angle OBC + \angle OCB + \angle BOC = 180^o$

$\angle OBC + \angle OCB = 180-60 = 120^o$

$2 \angle OBC = 120^o \Rightarrow \angle OBC = 60^o$

Therefore $\angle OBC = \angle OCB = \angle BOC = 60^o$

Therefore $\triangle BOC$ is equilateral

$\Rightarrow BC = OB = OC$

Hence $BC$ is equal to the radius of the circle.

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