MATHS 28-31
2007-06-11 12:41 am
回答 (2)
2007-06-11 1:06 am
✔ 最佳答案
28)CD=h/√3mBD=h/√3+100m
angleA=45 degrees
therefore,h=h/√3+100
h=50(3+√3)m
29a)cosA+cosB+sinA+sinB/sinA+sinB(cosAcos(90-A)=sinAsin(90-A))
=1+1=2
30)horizontal distance =√64^2-25^2
=√3471
=58.915....m
b)25/58.91
=1/2.35661
n=2.35661
31)tanβ=5/7
β=35.53767779
the reduced bearing of C from A =S54.46W
參考: me
收錄日期: 2021-04-13 00:50:13
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070610000051KK03218