Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
Let the side of the ascribed regular pentagon be 2 units.
(a) Calculating alpha. Let it be x.
a² = 5-4cos108
a = 2.49721204096 units.
(2.49721204096/sin108) = (2/sinb)
b = 49.61382244056°
c = 180-108-49.61382244056
c = 22.38617755944°
d² = 2.49721204096²+1-2*2.49721204096cos22.38617755944
d = 1.61803398876 units.
(1.61803398876/sin22.38617755944) = (2.49721204096/sine)
e = 180-36
e = 144°
It implies alpha, x is;
360-108-144-60
x = 48°
(b) Calculating AB/BC.
Notice;
BC = 1 unit.
Let AB be f.
(1/sin72) = (f/sin48)
f = AB = 0.78138871113 units.
It implies;
AB ÷ BC to 3 decimal places is;
0.78138871113 ÷ 1
= 0.78138871113
≈ 0.781
(c) Calculating Area Red ÷ Area Square.
Area Square is;
1²
= 1 square unit.
g² = 8-8cos108
g = 3.2360679775 units.
h² = g²-1²
h =3.07768353718 units.
i = √(2-2cos108)
i = 1.61803398875 units.
j² = i²-0.5²
j = 1.53884176859 units.
k = √(1²-0.5²)
k = 0.86602540378 unit.
l = h-j-k
l = 0.67281636481 unit.
m² = l²+1
m = 1.20527252551 units.
n = atan(1/0.67281636481)
n = 56.06668851812°
o = 360-180-120-56.06668851812
o = 3.93331148188°
p² = 1.20527252551²+1-2*1.20527252551cos3.93331148188
p = 0.21866570004 unit.
(0.21866570004/sin3.93331148188) = (1/sinq)
q = 18.28232358763°
r = 180-18.28232358763-3.93331148188
r = 157.78436493049°
s = 108-18.28232358763-atan(0.67281636481)
s = 55.78436493049°
t = 360-90-157.78436493049
t = 112.21563506951°
u = 180- 112.21563506951-55.78436493049
u = 12°
(v/sin55.78436493049) = (0.21866570004/sin12)
v = 0.86969908072 unit.
It implies;
Area red is;
Area triangle with height 0.86969908072 unit and base 0.21866570004sin112.21563506951 units.
= 0.5*0.86969908072*0.21866570004sin112.21563506951
= 0.0880281519 square units.
Therefore;
Area Red ÷ Area Square to 3 decimal places is;
0.0880281519 ÷ 1
= 0.0880281519
≈ 0.088
It implies;
(a) is 48°
(b) is 0.781
(c) is 0.088
