✔ 最佳答案
圖片參考：
http://us.f6.yahoofs.com/hkblog/Jnt8c_GBHw50sqWAbaC5UJE-_1/blog/20090511010340609.jpg.jpg?ib_____DiNPlj_1v
在 △HGI 中 ,由餘弦定理 :
(cos 30) (2HG * IG) = HG^2 + IG^2 - HI^2......★
設想 △FBC 向左平移 ,直至 C 與 G 重合 , 並記 F 的相應新位置為f ; B 的新位置為 b 。
在直角△HbG中, (bG = BC) / HG = cos(90 - 80),
得 HG = BC / cos 10......(1)
Hb / (bG = BC) = tan 10 , 得 Hb = BC * tan 10......(2)
在直角△IfG中, (fG = FC) / IG = cos(100 - 90),
得 IG = FC / cos 10......(3)
If / (fG = FC) = tan 10 , 得 If = FC * tan 10......(4)
******************************************************
HI^2 = (Hb + If)^2 + FB^2 (畢氏定理) , 代入 (2) , (4) :
HI^2 = (BC * tan 10 + FC * tan 10)^2 + FB^2
HI^2 = (tan 10)^2 *(BC^2 + FC^2 + 2BC * FC) + (FC^2 - BC^2)......(5)
代 (1), (3), (5) 入 ★ :
(2cos 30) (BC / cos 10)(FC / cos 10)
= (BC / cos 10)^2 + (FC / cos 10)^2 - (tan 10)^2 *(BC^2 + FC^2 +
2BC * FC) - (FC^2 - BC^2)
等式兩方同乘 (cos 10)^2 :
(2cos 30) BC * FC
= BC^2 + FC^2 - (sin 10)^2 *(BC^2 + FC^2 + 2BC * FC) -
(cos 10)^2 *(FC^2 - BC^2)
(2cos 30) BC * FC
= [1 - (sin10)^2 + (cos10)^2]*BC^2 + [1 - (sin10)^2 - (cos10)^2]*FC^2
- 2(sin10)^2 *BC * FC
(2cos 30) BC * FC
= 2(cos10)^2 *BC^2 - 2(sin10)^2 *BC*FC
等式兩方同除以 2BC*FC :
cos30 = (cos10)^2 *(BC/FC) - (sin10)^2
BC/FC = 0.9240423......
所以 FB/BC = √[( 1 - 0.9240423...^2) / (0.9240423...^2)]
= 0.41371484...
設所求的角為 x ,則 tan x = fH/HG = FB/HG, 由(1) :
= FB / (BC/cos10) = (FB/BC) * cos10
= 0.41371484... * cos10
x = 22.1674355... = 22.2 (3 s.f.)
2009-05-19 04:51:51 補充：
△FBC=△JEC
2009-05-19 04:54:25 補充：
F = J
B = E
