有關畢氏定理的麻煩問題【lop】
2013-05-20 3:31 am
圖片參考：http://imgcld.yimg.com/8/n/HA01076848/o/20130519192903.jpg
圖中 AB⊥BC 及 BC⊥DC ，AC=DE 和 BE=EC，若 AB=5，BC=12，求 AF 和 FC 的長。
回答 (5)
2013-05-20 7:47 am
2013-05-20 6:46 am
By Pythagoras thm., AC^2 = AB^2 + BC^2 = 5^2 + 12^2 = 169. so AC = 13 = DE.
DC^2 = DE^2 - EC^2 = 169 - 36 = 133, DC = 11.533.
Angle ACB = arctan (5/12) = 22.62 degree.
Angle ACD = 90 - 22.62 = 67.38 degree.
Angle EDC = arcsin (EC/DE) = arcsin(6/13) = 27.486 degree.
so angle DFC = 180 - 67.38 - 27.486 = 85.134 degree.
Applying sine rule to triangle DFC
FC/sin 27.486 = DC/sin 85.134 = 11.533/sin 85.134
so FC = 11.533 sin 27.486/sin 85.134 = 5.342.
AF = AC - FC = 13 - 5.342 = 7.658.
DC^2 = DE^2 - EC^2 = 169 - 36 = 133, DC = 11.533.
Angle ACB = arctan (5/12) = 22.62 degree.
Angle ACD = 90 - 22.62 = 67.38 degree.
Angle EDC = arcsin (EC/DE) = arcsin(6/13) = 27.486 degree.
so angle DFC = 180 - 67.38 - 27.486 = 85.134 degree.
Applying sine rule to triangle DFC
FC/sin 27.486 = DC/sin 85.134 = 11.533/sin 85.134
so FC = 11.533 sin 27.486/sin 85.134 = 5.342.
AF = AC - FC = 13 - 5.342 = 7.658.
2013-05-20 6:36 am
Actually do you know the answer?
2013-05-20 6:10 am
Re ✡ HYPERCUBE ✡,
sure
2013-05-19 22:39:06 補充：
知道又好，不知道又好，都是想拿一個方法而已。
sure
2013-05-19 22:39:06 補充：
知道又好，不知道又好，都是想拿一個方法而已。
2013-05-20 5:31 am
I'll use trigonometric function.
收錄日期: 2021-04-13 19:28:52
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130519000051KK00258