一隻船向北行駛10km，再沿方位角060度行駛了8km到達目的地，求出發點到目的地的距離和真方位角。(答案取自小數點後1個位)
我畫出黎5係直角3角形..所以5識得計..
thx for help..
答案如下..請列式
目的地的距離15.6km和真方位角026.3。
3角學應用的問題..sin cos tan哥d
2008-03-26 12:09 am
回答 (2)
2008-03-26 1:05 am
✔ 最佳答案
Assume the (0, 0) be the original departure point, then the 1st section is (0, 10), let the destination is (a, b)
a/8 = cos 30 = sq root3/2
a = 4 x sq root 3 = 6.9
(b-10)/a =tan 30=1/sqroot 3
(b-10)/(4 x sq root 3) = 1/sqroot 3
b=14
So the distance = sq root (6.9^2 + 14^)=15.6km
tan 真方位角 = 6.9/14
真方位角 = 26.3
2008-03-26 12:49 am
First I name the starting point to be A
after 向北行駛10km, the point be B
目的地be C
extend AB to D, such that triangle ADC is right triangle,
angle ADC = 90 degree
BD =8 cos60 =8*0.5 = 4
DC = 8sin60=8*(開方3)/2=4*(開方3)
so AC^2 = 14^2 + 4^2 *3 (Pyth. Thm.)
AC^2 = 196 + 48
AC^2 = 244
AC = 開方244 = 15.6 km
真方位角 = arc tan [4*(開方3) /14]
=026.3
after 向北行駛10km, the point be B
目的地be C
extend AB to D, such that triangle ADC is right triangle,
angle ADC = 90 degree
BD =8 cos60 =8*0.5 = 4
DC = 8sin60=8*(開方3)/2=4*(開方3)
so AC^2 = 14^2 + 4^2 *3 (Pyth. Thm.)
AC^2 = 196 + 48
AC^2 = 244
AC = 開方244 = 15.6 km
真方位角 = arc tan [4*(開方3) /14]
=026.3
參考: my math knowledge
收錄日期: 2021-04-13 15:20:27
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