Mathematics Question and Solution
a² = 12²+6²
a = 6√(5) cm.
a = AE.
b = atan(1/2)°
b = angle DAE.
c = 180-atan(2)
c = 116.56505117708°
c = angle AFC.
d = 180-116.56505117708-atan(1/2)
d = 36.86989764584°
(6/sin36.86989764584) = (e/sin116.56505117708)
e = 8.94427191 cm.
f = 6√(5)-8.94427191
f = 4.472135955 cm.
g = 180-2atan(2)
g = 53.13010235416°
g = angle AEB.
cos53.13010235416 = h/4.472135955
h = 2.683281573 cm.
It implies;
Area S1 is;
0.5*2.683281573*4.472135955sin53.13010235416
= 4.8 cm²
Calculating S2 Area.
j = 360-90-2atan(2)
j = 143.13010235416°
sin53.13010235416 = k/4.472135955
k = 3.577708764 cm.
l = 116.56505117708-90
l = 26.56505117708°
(m/sin(atan(1/2))) = (6/sin36.86989764584)
m = 4.472135955 cm.
n² = 4.472135955²+12²-2*4.472135955*12cos26.56505117708
n = 8.24621125124 cm.
(8.24621125124/sin26.56505117708) = (4.472135955/sino)
o = 14.03624346793°
p = 180-atan(2)
p = 116.56505117708°
q = 180-116.56505117708-14.03624346793
q = 49.39870535499°
tan49.39870535499 = 3.577708764/r
r = 3.066607512 cm.
S2 Area is;
0.5*3.066607512*3.577708764
= 5.48571428572 cm²
S1 ÷ S2 is;
4.8÷5.48571428572
= 0.875
= ⅞
