如图,AE与BC相交于点D,BD=CD,AD=ED,CA⊥AE.∠1=∠30°,且AB=3厘米,那么线段AC多长?别用勾
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如图,AE与BC相交于点D,BD=CD,AD=ED,CA⊥AE.∠1=∠30°,且AB=3厘米,那么线段AC多长?别用勾股定理,
![如图,AE与BC相交于点D,BD=CD,AD=ED,CA⊥AE.∠1=∠30°,且AB=3厘米,那么线段AC多长?别用勾](/uploads/image/z/17622026-26-6.jpg?t=%E5%A6%82%E5%9B%BE%2CAE%E4%B8%8EBC%E7%9B%B8%E4%BA%A4%E4%BA%8E%E7%82%B9D%2CBD%3DCD%2CAD%3DED%2CCA%E2%8A%A5AE.%E2%88%A01%3D%E2%88%A030%C2%B0%2C%E4%B8%94AB%3D3%E5%8E%98%E7%B1%B3%2C%E9%82%A3%E4%B9%88%E7%BA%BF%E6%AE%B5AC%E5%A4%9A%E9%95%BF%3F%E5%88%AB%E7%94%A8%E5%8B%BE)
因为AD=CD,BD=ED所以ABCE是平行形,AE=BC,角1=30度=ABD=角BAC,在直角三角形中30度角所对的直角边等于斜边的一半,得AE=BC=0.5*AB=1.5
如图,AE与BC相交于点D,BD=CD,AD=ED.CA⊥AE,∠1=30°,AB=3cm,那么线段AC多长
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