→ wayne2011 : 幾年前c大問的題目,參考黃家禮所編著的"幾明"解法2. 07/28 16:13
※ 編輯: wayne2011 (61.58.103.35), 07/28/2017 16:17:51
※ 編輯: wayne2011 (61.58.103.35), 07/28/2017 16:24:07
※ 引述《tcbt32 (唐)》之銘言:
: 10
: http://imgur.com/9RxAbwf
: 麻煩各位大大了
三角形MBA與BMN中
BM/sin(4pi/9)=AB/sin(2pi/9)...(1)
BM/sin[(8pi/9)-BMN]=BN/sinBMN...(2)
(1)/(2):sin[(8pi/9)-BMN]/sin(4pi/9)=sinBMN/sin(2pi/9)...ABN為其"等腰"三角
sin[(pi/9)+BMN]=2cos(2pi/9)*sinBMN
sin(pi/9)*cosBMN=sin(BMN)*[2cos(2pi/9)-cos(pi/9)]
cotBMN=[2sin(5pi/18)-cos(pi/9)]/sin(pi/9)
={2sin[(pi/9)+(pi/6)]-cos(pi/9)}/sin(pi/9)
=[sqrt(3)sin(pi/9)+cos(pi/9)-cos(pi/9)]/sin(pi/9)
=sqrt(3)...因而,BMN=pi/6.
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