特別角的三角函數(1)
Author: 藍邦偉  老師
Date: 25 12 月 2012

$1$

$2$

$2$

$ 60° $

$ A $

$ 2 $

$ C $

$ D $

$ B $

$ 30° $

$ \sqrt{3} $

$ B' $

$ C' $

$ 1 $

$ \left\lbrace\begin{array}{ll} sin30°=\frac{1}{2} \ cos30°=\frac{\sqrt{3}}{2} \ tan30°=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3} \end{array}   \right.\  $

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特別角30°

u = Vector((x(Corner[1]), 0), (x(Corner[3]), 0))
v = Vector((0, y(Corner[1])), (0, y(Corner[3])))
A = (-1.149799576952746, 1.1988717460488123)
B: A + (sqrt(3), 0)
C: B + (0, 1)
a = Segment(A, B)
b = Segment(B, C)
c = Segment(C, A)
r = 2.06 (slider from 1 to 3, step 0.01)
B' = Dilate(B, r, A)
C' = Dilate(C, r, A)
f = Segment(B', C')
D = Mirror(C, B)
g = Segment(A, D)
h = Segment(B, D)
k = Circle(A, C, D)
s = 0 (slider from 0 to 6, step 1)
α = Angle(C, B, A)
d = Segment(C, C')
e = Segment(B, B')
poly1, c_1, a_1, b_1 = Polygon(A, B, C)