tag:blogger.com,1999:blog-3550204829234352390.post536383341857020865..comments2024-03-27T09:13:48.546+08:00Comments on 研發養成所 ( Bridan's Blog - 4rdp, For R&D Person ): 訓練數學感 153 ─ 角度 BAE 有多大Bridanhttp://www.blogger.com/profile/17055047757114667099noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-3550204829234352390.post-84262542477817796102017-11-18T23:22:32.861+08:002017-11-18T23:22:32.861+08:00謝謝西瓜幫忙解答。謝謝西瓜幫忙解答。Bridanhttps://www.blogger.com/profile/17055047757114667099noreply@blogger.comtag:blogger.com,1999:blog-3550204829234352390.post-45634368289458874832017-11-18T22:57:23.334+08:002017-11-18T22:57:23.334+08:00DE=DA,是故∠AED=∠EAD
由外角定理,50°= ∠EDC = ∠AED + ∠EAD = ...DE=DA,是故∠AED=∠EAD<br />由外角定理,50°= ∠EDC = ∠AED + ∠EAD = 2∠AED = 2∠EAD = 2 x 25°.https://www.blogger.com/profile/16677804023065232981noreply@blogger.comtag:blogger.com,1999:blog-3550204829234352390.post-72321329368057328682017-11-18T18:51:03.391+08:002017-11-18T18:51:03.391+08:00為什麼∠EAC=50°/2
為什麼∠EAC=50°/2<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3550204829234352390.post-4240277897143168352017-11-17T23:46:34.938+08:002017-11-17T23:46:34.938+08:00正解,謝謝提供這麼棒的解法,有興趣的朋友可以留言其它解法。正解,謝謝提供這麼棒的解法,有興趣的朋友可以留言其它解法。Bridanhttps://www.blogger.com/profile/17055047757114667099noreply@blogger.comtag:blogger.com,1999:blog-3550204829234352390.post-61527284331990639322017-11-17T22:36:58.924+08:002017-11-17T22:36:58.924+08:00想了很多種方法,應該都是等價的。下面是效率最高的一個。
※引理:兩切線的夾角與所夾劣弧度數互補
...想了很多種方法,應該都是等價的。下面是效率最高的一個。<br /><br />※引理:兩切線的夾角與所夾劣弧度數互補<br /><br /><br />所求=∠BAC-∠EAC<br />=(180°-BC弧)-(50°/2)<br />=[180°-(EC弧+20°)]-25°<br />=[180°-(180°-∠EDC+20°)]-25°<br />=[180°-(180°-50°+20°)]-25°<br />爽快地把180度消掉,計算得5°。.https://www.blogger.com/profile/16677804023065232981noreply@blogger.com