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प्रश्न
In the given figure, an equilateral triangle ABC is inscribed in a circle with center O.
Find: (i) ∠BOC
(ii) ∠OBC
उत्तर
In the given figure, ABC is an equilateral triangle.
Hence all the three angles of the triangle will be equal to 60°
i.e. ∠A = ∠B = ∠C = 60°
As the triangle is an equilateral triangle, BO and CO will be the angle bisectors of B and C respectively.
Hence ∠OBC = `"∠ABC"/2`
= 30°
and as given in the figure we can see that OB and OC are the radii of the given circle.
Hence they are of equal length.
The ΔOBC is an isosceles triangle with OB = OC
In ΔOBC,
∠OBC = ∠OCB as they are angles opposite to the two equal sides of an isosceles triangle.
Hence, ∠OBC = 30° and ∠OCB = 30°
Since the sum of all angles of a triangle is 180°
Hence in triangle OBC, ∠OCB + ∠OBC + ∠BOC + BOC = 180°
30° + 30° + ∠BOC= 180°
60° + BOC = 180°
∠BOC = 180° - 60°
∠BOC = 120°
Hence ∠BOC =120° and ∠OBC =30°
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