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maths (20 分)
回答 (3)
2010-07-22 1:13 am
2010-07-21 5:58 am
已知tanA=3/4,計算 3sinA-4cosA/3sinA+4cosA
Ans:
tanA=3/4
斜邊 = √(3^2 + 4^2) = 5 (畢氏定律)
鄰邊 = 4
對邊 = 3
由此, sinA = 3/5 , cos = 4/5
3sinA - 4cosA/3sinA + 4cosA
=3sinA – 4/3(tanA) + 4 cos A
=3(3/5) – 4/3(3/4) + 4(4/5)
=9/5 – 16/9 + 16/5
=(9/5)(9/9) – (16/9)(5/5) + (16/5)(9/9)
=81/45 – 80/45 + 144/45
=(81-80+144) / 45
=29/9
Ans:
tanA=3/4
斜邊 = √(3^2 + 4^2) = 5 (畢氏定律)
鄰邊 = 4
對邊 = 3
由此, sinA = 3/5 , cos = 4/5
3sinA - 4cosA/3sinA + 4cosA
=3sinA – 4/3(tanA) + 4 cos A
=3(3/5) – 4/3(3/4) + 4(4/5)
=9/5 – 16/9 + 16/5
=(9/5)(9/9) – (16/9)(5/5) + (16/5)(9/9)
=81/45 – 80/45 + 144/45
=(81-80+144) / 45
=29/9
參考: try rather than know everything
2010-07-21 4:04 am
已知tanA=3/4,計算 3sinA-4cosA/3sinA+4cosA
tanA=3/4
A=36.86989765
3sinA-4cosA/3sinA+4cosA
=3sin36.86989765-4cos36.86989765/3sin36.86989765+4cos36.86989765
=3.22(corr.to 3 sig.fig.)
Hope you will understand!!! If you have any question, you may ask me through email!
Thanks!!!!!!
tanA=3/4
A=36.86989765
3sinA-4cosA/3sinA+4cosA
=3sin36.86989765-4cos36.86989765/3sin36.86989765+4cos36.86989765
=3.22(corr.to 3 sig.fig.)
Hope you will understand!!! If you have any question, you may ask me through email!
Thanks!!!!!!
收錄日期: 2021-04-23 20:42:58
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100720000051KK01431