搵3角形中的長度

AF = 4/cos 15 = 4.1411
AB = 2/cos 30 = 2.3094
By Cosine Rule, BF^2 = AF^2 + AB^2 - 2AB . AF. cos 45
= 8.95726, so BF = sqrt (8.95726) = 2.9929
By sine rule,
AB/sin (angle AFB) = BF/sin 45
2.3094/sin(angle AFB) = 2.9929/sin 45
Angle AFB = arcsin ( 2.3094 sin 45/2.9929) = arcsin ( 0.545622) = 33.067 degree.
Let AX be the green line,
so angle ABX = 45 + 33.067 = 78.067 degree ( ext. angle of triangle)
so length of green line = AX = AB sin 78.067 = 2.259m.

let the length of green line be L,
let angle AFB be X degree,

AF=4/cos15
AB=2/cos30

BF/cos45=AB/cosX=2/(cos30·cosX)
BF=2·cos45/(cos30·cosX)
cosX=2·BF·cos45/cos30 =L/AF

2·BF·cos45/cos30 =L/AF
L=2·BF·AF·cos45/cos30 = BF·cos15·cos45/(2·cos30)
跟住你就自行化簡.


2010-11-01 16:48:45 補充：
SORRY, 睇錯, 我以為要in term of BF