题目内容：

连接直角三角形ABC的直角顶点C与斜边AB的两个三等分点DE 所得线段CD CE的长分别为sina和cosa
（0＜a＜π／2）,则AB长为

优质解答

如图,设直角三角形边长AC=x,BC=y,AD=DE=EB=z    则有AB=3z
由余弦定理可知：CD^2=AD^2+AC^2-2AD*AC*cosA=(sina)^2          式1  cosA=x/3z  (邻边比斜边)
                            CE^2=EB^2+BC^2-2EB*BC*cosB=(cosa)^2          式2  cosB=y/3z(邻边比斜边)
上两等式相加,代入数据则有：2z^2+x^2+y^2-2x^2/3-2y^2/3=1
                               再次化简：（x^2+y^2)/3+2z^2=1       根据勾股定理又知：x^2+y^2=AB^2=9z^2
                                                             3z^2+2z^2=1 
解得                                                                    z=(根号5)/5
所以AB=3z=3*(根号5)/5