三角的簡易應用 (1)
2011-03-26 10:22 pm
回答 (1)
2011-03-27 12:27 am
✔ 最佳答案
1.draw a st. line A'E',such that 垂直於BEsin30=AE/38
38．sin30=AE
AE=19m
sin70=A'E'/38
38．sin70=A'E'
A'E'=35.71m(corr to 4sig.fig.)
鉛垂距離是=35.71-19
=16.7m(corr to 3sig.fig.)
2.let height of the tree=BC,
tan25=BC/16
16．tan25=BC
BC=7.46(corr to 3sig.fig.)
height of the tree is 7.46m
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https://hk.answers.yahoo.com/question/index?qid=20110326000051KK00726