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⊥

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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⊥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,……,则第10条线段A5C5等于多少?
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解法一：利用相似求解
AC=3,BC=4,Rt△ABC中,AB=5
显然：△A1CA∽△CBA,
得A1C/CB=CA/BA,得A1C=(CA/BA)*CB=3×(4/5)
显然：△C1A1C∽△A1CA,
得C1A1/A1C=A1C/CA,得C1A1=(A1C)^2/CA=3×(4/5)^2
显然：△A2C1A1∽△C1A1C,
得A2C1/C1A1=C1A1/A1C,得A2C1=(C1A1)^2/A1C=3×(4/5)^3
显然：△C2A2C1∽△A2C1A1,
得C2A2/A2C1=A2C1/A1C1,得C2A2=(A2C1)^2/C1A1=3×(4/5)^4
同理可得：
△C3A3C2∽△A3C2A2,得C3A3=3×(4/5)^6
△C4A4C3∽△A4C3A3,得C4A4=3×(4/5)^8
△C5A5C4∽△A5C4A4,得C5A5=3×(4/5)^10
所求C5A5=3×4^10/5^10=3145728/9765625=0.3221225472
解法二：利用三角函数求解
令∠B=α,得cosα=BC/AB=4/5
所以A1C=ACcosα=3×(4/5)
A1C1=A1Ccosα=3×(4/5)^2
C1A2=A1C1cosα=3×(4/5)^3
A2C2=C1A2cosα=3×(4/5)^4
.
A5C5=3×(4/5)^10