特別角的三角函數(3)
Author: 藍邦偉  老師
Date: 26 12 月 2012

$O$

$2$

$Step=%v$

$ \left\lbrace \begin{array}{ll} sin60°=\frac{\sqrt{3}}{2} \ cos60°=\frac{1}{2} \ tan60°=\frac{\sqrt{3}}{1}=\sqrt{3} \end{array}  \right.\ $

$ A $

$ 60° $

$ C $

$ B $

$ 2 $

$ 2 $

$ 1 $

$ M $

$ \sqrt{3} $

$ B' $

$ M' $

特別角60°

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t = 0.39 (slider from 0 to 3.3, step 0.01)
O: 0.5 * (Corner[1] + Corner[3])
r: 0.4 * (y(Corner[4]) - y(Corner[1]))
B: O + (0, r)
A = Rotate(B, 120*°, O)
E: A + (t, 0)
c = Circle(O, r)
C = Rotate(B, 240*°, O)
a = Segment(A, C)
b = Segment(C, B)
d = Segment(B, A)
M = Midpoint(A, C)
e = Segment(B, M)
n = 0 (slider from 0 to 5, step 1)
f = Segment(A, M)
g: x = t
D: (x(E), sqrt(3) * t + y(E))
多邊形1, e_1, a_1, d_1 = Polygon(A, D, E)
α = Angle(A, M, B)
h = Segment(D, E)