#### 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°

⇒ 2OBC = 180° - 60°

⇒ 2∠OBC = 120°

⇒ `OBC = (120°)/2 = 60°`

⇒ ∠OBC = ∠OCB = ∠BOC = 60°

⇒ ΔBOC is an equilateral triangle

⇒ BC = OB = OC

But, OB and OC are the radii of the circum – circle

∴ BC is also the radius of the circum – circle

Is there an error in this question or solution?

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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. Concept: Arc and Chord Properties - If Two Arcs Subtend Equal Angles at the Center, They Are Equal, and Its Converse.

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