MB是△MAC的中线,AB = BC = 1 ,∠AMB = 45°,∠BMC = 30° ,求M到BC的距离.
过点A作AD⊥MB于D,过点C作CE⊥MB于E,过点M作MH⊥AB于H.设 AD = x ；
则有：MD = x ,CE = x ,ME = √3x ,DE = (√3-1)x ,BD = (√3-1)x/2 ,MB = (√3+1)x/2 ；
由勾股定理可得：AD²+BD² = AB² ,则有：x² = 2(4+√3)/13 ；
因为,(1/2)*AB*MH = △ABM面积 = (1/2)*MB*AD ,
所以,MH = MB*AD/AB = (√3+1)x²/2 = (7+5√3)/13 ,
即：塔到直路ABC的最短距离为 (7+5√3)/13 千米.