IF tanA=m/n, then cosA=?
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數學有問題想請教
2010-05-25 10:47 pm
回答 (3)
2010-05-26 12:08 am
✔ 最佳答案
圖片參考：http://imgcld.yimg.com/8/n/HA00132885/o/701005250051513873414260.jpg
As shown in the figure above,
If tanA = m/n
Then cos A = n/SQRT(m^2 + n^2) or -n/SQRT(m^2 + n^2) [depend on the range of the angle A]
2010-05-25 11:14 pm
tanA=m/n
=>sinA/cosA=m/n
=>let cosA = k ,
then sinA = mk/n
As (sinx)^2+(cosx)^2=1
=>(k^2)+(mk/n)^2=1
=>k^2(n^2+m^2)=n^2
=>k=n/sqrt(m^2+n^2)//
so cosx=n/sqrt(n^2+m^2).
=>sinA/cosA=m/n
=>let cosA = k ,
then sinA = mk/n
As (sinx)^2+(cosx)^2=1
=>(k^2)+(mk/n)^2=1
=>k^2(n^2+m^2)=n^2
=>k=n/sqrt(m^2+n^2)//
so cosx=n/sqrt(n^2+m^2).
2010-05-25 11:06 pm
tan A = sinA/cosA
therefore sinA/cosA = m/n
therefore nsinA = mcosA
therefore cosA = (n sinA)/m or (n/m) sinA
hope this is right!
2010-05-25 23:30:09 補充：
the other two answer is a bit complicated .... isn't it?
therefore sinA/cosA = m/n
therefore nsinA = mcosA
therefore cosA = (n sinA)/m or (n/m) sinA
hope this is right!
2010-05-25 23:30:09 補充：
the other two answer is a bit complicated .... isn't it?
收錄日期: 2021-04-23 23:47:42
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100525000051KK00515