急求一道三角函数的应用题!A man can row at 16 km/h,and run at 12 km/h.He needs to get from a point A,on the south bank of a stretch of still water,to point B on the north bank of the water.The direct distance from A to B is 20 km,and the w
急求一道三角函数的应用题!A man can row at 16 km/h,and run at 12 km/h.He needs to get from a point A,on the south bank of a stretch of still water,to point B on the north bank of the water.The direct distance from A to B is 20 km,and the w
急求一道三角函数的应用题!
A man can row at 16 km/h,and run at 12 km/h.He needs to get from a point A,on the south bank of a stretch of still water,to point B on the north bank of the water.The direct distance from A to B is 20 km,and the water is 16 km wide.He starts rowing with an angle between North and the direction in which he rows.Find an expression for the time T he will take to get from A to B,in terms of .
说,一个男人划船的速度可以达到16公里每小时,跑步的速度可以达到12公里每小时,他需要从A点（南边河岸）出发,到达B点（河的北岸）.A到B点距离是20公里,河有16公里宽,他划船的方向角度是北向南θ度,求这个人从A点到B点需要的时间与这个角度的函数关系
不考虑河流的速度
T T
急求一道三角函数的应用题!A man can row at 16 km/h,and run at 12 km/h.He needs to get from a point A,on the south bank of a stretch of still water,to point B on the north bank of the water.The direct distance from A to B is 20 km,and the w
好像没有北向南θ度这种说法,“between North and the direction in which he rows”我理解是他划船的角度是从北向B点方向偏θ度
划船时间为：16/cosθ/16=1/cosθ
跑步时间为：|（20^2-16^2）^0.5-16*tgθ|/12=|1-4/3tgθ|
t=1/cosθ+|1-4/3tgθ|
有绝对值是因为当θ的角度大于AB直线与正北方向夹角时,其值为负数
设他从A到B的运动角度是a,
则他到达的时间T=16/(16cosa)+[12-16(tan a)]/12
设时间为t秒： t=（12-16×tanθ）/12+（16×cosθ）/16 =1-4/3×tanθ+cosθ