Mathematics Question and Solution
Calculating angle EHB.
Let the side length of the regular pentagon be 1 unit.
Therefore;
AB = BC = CD = DE = AE = 1 unit.
a² = 2-2cos108
a = 1.6180339887 units.
a is BE, the square side length.
b = 108-½(180-108)
b = 108-½(72)
b =108-36
b = 72°
b is angle BED
c = 90-b
c = 90-72
c = 18°
c is angle DEG.
sin18 = d/1
d = 1sin18
d = 0.3090169944 units.
cos18 = e/1
e = 1cos18
e = 0.9510565163 units.
f = a-e
f = 1.6180339887-0.9510565163
f = 0.6669774724 units.
tang = 0.3090169944/0.6669774724
g = 24.8587372091°
g is angle EGH.
h = 90-g
h = 90-24.8587372091
h = 65.1412627909°
h is angle FGH.
cos65.1412627909 = (0.5*1.6180339887)/j
j = 1.9244783946 units.
j is GH.
k² = 1.9244783946²+1.6180339887²-2*1.9244783946*1.6180339887cos24.8587372091
k = 0.8191014929 units.
Therefore, angle EHB is;
Let it be m.
sinl = (0.5*1.6180339887)/0.8191014929
l = 81°
It implies;
m = 2l
m = 2*81
m = 162°
Again, m is angle EHB.
