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本帖最後由 L.K. 於 1-11-2009 13:46 編輯某壇見到條數..幾好玩.
圖是正方形，問角多大
本帖最後由 -終場ソ使者- 於 1-11-2009 10:16 編輯
Let x be the length of the square
and α,β,γ be the angle between length √3 and 1,1 and √5,√3 and √5,respectively.
By cosine law,
α=cos-1[(3+1-x2)/2√3]
β=cos-1[(5+1-x2)/2√5]
γ=cos-1[(3+5-2x2)/2√15]
α+β+γ=180'
=>cos-1[(3+1-x2)/2√3]+cos-1[(5+1-x2)/2√5]+cos-1[(3+5-2x2)/2√15]=180'
x=...
then,calculate α 本帖最後由 【YU】 於 1-11-2009 12:21 編輯
2# -終場ソ使者-
α+β+γ=180' <= should be 360'
cos-1[(3+1-x2)/2√3]+cos-1[(5+1-x2)/2√5]+cos-1[(3+5-2x2)/2√15]=360'
^ need numerical method http://nakuz.com/bbs/images/smilies/Tuzki/tk_24.gif 係喎
個x點搵 唔好諗得咁複雜 新鮮滾熱辣計到....
let a be the length of the square , theta be the angle between a and 1 , gamma be the angle between root3 and 1
0<θ<90 , 0<γ<180(跟圖架咋)
3 = 1 + a^2 - 2acosθ
cosθ= -(2 - a^2) / (2a)
5 = 1 + a^2 - 2asinθ
sinθ = -(4 - a^2) / (2a)
∴ [(2 - a^2)^2 + (4 - a^2)^2] / (4a^2) = 1
a^4 - 8a^2 + 10 = 0
∴ a^2 = 4 + root6 or 4 - root6(rej.) <----至於點解自己check一check就ok
∴ a^2 = 3 + 1 - 2root3cosγ
4 + root6 = 4 - 2root3cosγ
cosγ = - root2 / 2
γ = 135°
∴The angle is 135°.
用中3程度GE方法 用中3程度GE方法
L.K. 發表於 2-11-2009 18:26 http://www.nakuz.com/bbs/images/common/back.gif
會唔會用得上similar triangle?
會唔會用得上similar triangle?
Roy 發表於 2/11/2009 11:08 PM http://nakuz.com/bbs/images/common/back.gif
唔會.. ...
係咪要好似砌puzzle咁砌出黎
之後3個數都用唔著?