在AC上取一点D,连接BD,使得:∠BDC = 30° ;
在Rt△BCD中,∠C = 90°,∠BDC = 30° ,BC = 1 ,
可得:BD = 2BC = 2 ,CD = √3BC = √3 ;
因为,∠ABD = ∠BDC-∠A = 15° = ∠A ,
所以,AD = BD = 2 ;
可得:AC = AD+CD = 2+√3 ,
所以,tan15° = tan∠A = BC/AC = 1/(2+√3) = 2-√3 .
在AC上取一点D,连接BD,使得:∠BDC = 30° ;
在Rt△BCD中,∠C = 90°,∠BDC = 30° ,BC = 1 ,
可得:BD = 2BC = 2 ,CD = √3BC = √3 ;
因为,∠ABD = ∠BDC-∠A = 15° = ∠A ,
所以,AD = BD = 2 ;
可得:AC = AD+CD = 2+√3 ,
所以,tan15° = tan∠A = BC/AC = 1/(2+√3) = 2-√3 .