19/08/13 00:57:34.59 gccQR1zi.net
>>657
AD=BC=1 とすると
A (0, 0)
B (b, 0)
C (c, sin(B)) = (c, 0.9135454576426)
D (cos(A), sin(A)) = (-(√3)/2, 1/2)
ただし
b = (1/2)√(5+2√5) + cos(A) = 0.672816364803
c = b - cos(B) = 1.079553007879
これより
tan(∠ABD) = sin(A)/{b-cos(A)} = 1/√(5+2√5) = tan(18ﾟ)
∴ ∠ABD = 18ﾟ >>658