如图.已知Rt△ABC中,AC=3,BC=4,过直角顶点C作CA1⊥AB,垂足为A1,再过A1作A1C1⊥BC,垂足为C
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如图.已知Rt△ABC中,AC=3,BC=4,过直角顶点C作CA1⊥AB,垂足为A1,再过A1作A1C1⊥BC,垂足为C1,过C1作C1A2⊥AB,垂足为A2,再过A2作A2C2⊥BC,垂足为C2,…,这样一直做下去,得到了一组线段CA1,A1C1,C1A2,…,则CA1=______,
C
∵Rt△ABC中,AC=3,BC=4,
∴AB= AC2+BC2=5, ∵CA1⊥AB, ∴CA1= AC•BC AB= 12 5,cos∠B= AC AB= 4 5, ∵A1C1⊥BC, ∴∠CA1B=∠A1C1B=90°, ∴∠CA1C1+∠A1CB=∠B+∠A1CB=90°, ∴∠CA1C1=∠B, 同理:∠AnCnAn+1=∠B, ∴cos∠AnCnAn+1= CnAn+1 AnCn= 4 5. 故答案为: 12 5, 4 5.
如图.已知Rt△ABC中,AC=3,BC=4,过直角顶点C作CA1⊥AB,垂足为A1,再过A1作A1C1⊥BC,垂足为C
如图,已知Rt△ABC中,AC=3,BC=4,过直角顶点C作CA1⊥AB,垂足为A1,再过A1作A1C1⊥BC,垂足为C
如图,已知Rt△ABC中,AC=3,BC=4,过直角顶点C作CA1⊥AB,垂足为A1,再过A1作A1C1⊥BC,垂足为C
已知Rt△ABC中,AC=3,BC= 4,过直角顶点C作CA1⊥AB,垂足为A1,再过A1作A1C1⊥BC
如图,已知直角三角形ACB,AC=3,BC=4,过直角顶点C作CA1⊥AB,垂足为A1,再过A1作A1C1⊥BC,垂足为
已知Rt△ABC中,AC=3,BC=4,过直角顶点C作CA1⊥AB,垂足为A1,
一道代数题.如图,已知直角三角形ACB,AC=3,BC=4,过直角顶点C作CA1⊥AB,垂足为A1,再过A1作A1C1⊥
初二相似图形应用题如图,已知RT△ABC,AC=3,BC=4,通过直角顶点C作CA1⊥AB,垂足为A1;再过A1作A1C
已知Rt△ABC中,AC=3,BC= 4,过直角顶点C作CA1⊥AB,垂足为A...
如图,已知直角三角形ABC中,∠ACB=90°,AC=3,BC=4,过直角顶点C作CA1⊥ABm,垂足为A1,
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