BEC is an equilateral triangle in a square ABCD, find x
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OBC = 45° (caz' of diagonal segment)
BCO = 60° (Since BEC is an equilateral ∆)
So, In ∆OBC,
> OBC + BCO + x = 180°
> x = 180° - 60° - 45° = 75°
BCO = 60° (Since BEC is an equilateral ∆)
So, In ∆OBC,
> OBC + BCO + x = 180°
> x = 180° - 60° - 45° = 75°
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