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ABC + CDA = ABCD ask what are ABCD In ABC, a is a hundred, B is a ten, C is a single, in CDA, C is a hundred, D is a ten, a is a single, in ABCD, a is a thousand, B is a hundred, C is a ten, D is a single
0 respectively,
Delta ABC is an equilateral triangle, O is a point in delta ABC, ∠ AOB = 110, ∠ BOC = 135. Calculate the degree of each internal angle of the triangle formed by line segments Ao, Bo and Co
[answer with rotation knowledge] rotate the triangle ABO by 60 degrees anticlockwise at point a to get the triangle ACD, so the triangle ABO is equal to the triangle ACD. You can also say that draw an angle of oad = 60 degrees on the right side of AC, make ad = Ao, connect DC, so the triangle ABO is equal to the triangle ACD (SAS) angle AOC = 115 degrees, angle ADC
In the triangle ABC, take any point O, connect Ao, Bo, Co, then the relationship between ∠ A and ∠ BOC is?
It is not necessary to connect Ao in the title, otherwise ∠ a in the title is not clear. If ∠ a ＜ BOC is the circumscribed circle of △ ABC, then ∠ A is the circular angle of the chord BC, and ∠ BOC is the inner angle of the chord BC, while in the same circle, the circular angle of the same chord is less than the inner angle of the same chord, and ∠ a ＜ BOC
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