Question

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⊥

Answer 1

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