點解sin60 = (開方3) / 2 ?

Draw a right- angled triangle ABC with angle A = 60 degree and angle B = 90 degree.
Draw M and N such that M and N are mid-point of AC and BC respectively.
Therefore triangle CMN congruent triangle triangle BMN (SAS)
so CM = MB = AM = x.
Since AM = MB and angle A = 60, therefore, triangle ABM is an equilateral triangle, that is AB = AM = MB = x.
Now AC = 2 AC = 2x
AB = x. By Pythagoras theorem, CB = sqrt [ (2x)^2 - x^2] = x sqrt 3.
Therefore sin 60 = CB/AC = x sqrt 3/ 2x = (sqrt 3)/2



2009-06-09 16:07:18 補充：
Correction : Line 7 should be AC = 2 AM = 2x.

其實(開方3) / 2 ,只係一個比例,
佢係指2條直角3角形既邊既關係,
當對邊除鄰邊係(開方3) / 2 ,
計緊既果隻角就係60度

這是根據那個特定三角形的比例來計算出來的。