Mathematics
In the given figure, ABC is a triangle in which ∠BAC = 30°. Show that BC is equal to the radius of the circumcircle of the triangle ABC, whose centre is O.

Circles
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Answer
Join OB and OC.

We know that,
Angle at the centre is twice the angle at remaining circumference.
∴ ∠BOC = 2∠BAC = 2 × 30° = 60°
In △OBC,
⇒ OB = OC [Radii of same circle]
⇒ ∠OBC = ∠OCB = x (let) [As angles opposite to equal sides are equal]
⇒ ∠BOC + ∠OBC + ∠OCB = 180°
⇒ 60° + x + x = 180°
⇒ 2x + 60° = 180°
⇒ 2x = 180° - 60°
⇒ 2x = 120°
⇒ x = = 60°.
∴ ∠OBC = ∠OCB = ∠BOC = 60°.
Hence, △OBC is an equilateral triangle.
∴ OB = OC = BC.
Hence, proved that BC is equal to the radius of the circumcircle of the triangle ABC.
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