圖片參考：http://photos-a.ak.fbcdn.net/hphotos-ak-snc1/hs034.snc1/2595_1115424127617_1287361866_341288_6336532_n.jpg
更新1:
I need a detail proof.
更新2:
I need a faster and simplier proof. (NOT COS FORMULA)
Maths (Hard)
2009-03-06 2:12 am
Please solve the following question
圖片參考：http://photos-a.ak.fbcdn.net/hphotos-ak-snc1/hs034.snc1/2595_1115424127617_1287361866_341288_6336532_n.jpg
圖片參考：http://photos-a.ak.fbcdn.net/hphotos-ak-snc1/hs034.snc1/2595_1115424127617_1287361866_341288_6336532_n.jpg
回答 (4)
2009-03-11 11:01 am
✔ 最佳答案
圖片參考：http://euler.tn.edu.tw/images/think045.gif
In the figure, AB = BC,ㄥABC = 90, ㄥ2 = b, ㄥ3 = c
soㄥ2+ㄥ3 = b + c = 45, it is very easy to know that a = 45
So a + b + c = 45 + 45 = 90
2009-03-07 7:48 am
2009-03-07 2:09 am
It is clear that a = 45* but the proof can still be completed without finding b and c. By considering tan b = 1/2 and tan c = 1/3,
tan ( b + c ) = ( tan b + tan c ) / ( 1 - tan b tan c )
= ( 1/2 + 1/3 ) / ( 1 - 1/6 )
= 1
As b + c < 90*, b + c = 45*
Lastly, a + b + c = 45* + 45* = 90*
tan ( b + c ) = ( tan b + tan c ) / ( 1 - tan b tan c )
= ( 1/2 + 1/3 ) / ( 1 - 1/6 )
= 1
As b + c < 90*, b + c = 45*
Lastly, a + b + c = 45* + 45* = 90*
2009-03-06 2:34 am
let the side length of each square is l
a=45
l^2=(2l)^2+[l^2+(2l)^2]-2(2l){[l^2+(2l)^2]^(1/2)}cos b
1=(2)^2+[1+(2)^2]-2(2)[5^(1/2)]cos b
1=4+5-4[5^(1/2)]cos b
b=26.56505118
l^2=(3l)^2+[l^2+(3l)^2]-2(3l){[l^2+(3l)^2]^(1/2)}cos c
1=(3)^2+[1+(3)^2]-2(3)[10^(1/2)]cos c
1=9+10-2(3)[10^(1/2)]cos c
c=18.43494882
So, a+b+c=90
a=45
l^2=(2l)^2+[l^2+(2l)^2]-2(2l){[l^2+(2l)^2]^(1/2)}cos b
1=(2)^2+[1+(2)^2]-2(2)[5^(1/2)]cos b
1=4+5-4[5^(1/2)]cos b
b=26.56505118
l^2=(3l)^2+[l^2+(3l)^2]-2(3l){[l^2+(3l)^2]^(1/2)}cos c
1=(3)^2+[1+(3)^2]-2(3)[10^(1/2)]cos c
1=9+10-2(3)[10^(1/2)]cos c
c=18.43494882
So, a+b+c=90
參考: 自己
收錄日期: 2021-04-21 22:02:11
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090305000051KK01077