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Question
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.
Solution
Given – In the figure ABC is a triangle in which ∠A = 30°
To prove – BC is the radius of circumcircle of ∆ABC whose centre is O.
Construction – Join OB and OC.
Proof – ∠BOC = 2∠BAC = 2 × 30° = 60°
Now in ∆OBC,
OB = OC ...[Radii of the same circle]
∠OBC = ∠OCB
But, in ΔBOC,
∠OBC + ∠OCB + ∠BOC = 180° ...[Angles of a triangle]
`=>` ∠OBC + ∠OBC + 60° = 180°
`=>` 2∠OBC + 60° = 180°
`=>` 2∠OBC = 180° – 60°
`=>` 2∠OBC = 120°
`=> ∠OBC = 120^circ/2 = 60^circ`
`=>` ∠OBC = ∠OCB = ∠BOC = 60°
`=>` ΔBOC is an equilateral triangle
`=>` BC = OB = OC
But, OB and OC are the radii of the circumcircle
∴ BC is also the radius of the circumcircle.
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