A,B,c are three points on circle the centre is zero and every point is equal distance others two points find angle AOB.Plz answer fastly
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Since A B C are equidistant from each other, then △ABC is an equilateral triangle
So all of its angles are 60°
Now, AO = BO = CO = radius of circle
In the triangles △AOB, △BOC, △COA, we have AO = BO, BO = CO, CO = AO
So, △AOB, △BOC, △COA are isosceles triangles,
But, in these three triangles, AB = BC = CA, so they are congruent to each other.
Therefore, ∠AOB = ∠BOC = ∠COA
Also, these three angles form a circle, so they must add up to 360
So, ∠AOB + ∠BOC + ∠COA = 360
=> 3(∠AOB) = 360
∠AOB = 120
Hope this helped!
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