Mathematics Question and Solution
a² = 3²+4²
a = 5 cm.
Where a is AC.
b = atan(4/3)°
Where b is angle DAC.
c = 60+atan(4/3)
c = 113.13010235416°
Where c is angle EAC.
d² = 3²+5²-2*3*5cos113.13010235416
d = 6.76643256752 cm.
Where d is CE.
(6.76643256752/sin113.13010235416) = (3/sine)
e = 24.06202137986°
Where e is angle ACF.
f = atan(3/4)-24.06202137986
f = 12.80787626598 cm.
Where f is angle DCE.
tan(12.80787626598) = g/4
g = 0.90935597101 cm.
Where g is DF.
cos12.80787626598 = 4/h
h = 4.10206390516 cm.
Where h is CF.
i = 3-0.90935597101
i = 2.09064402899 cm.
Where i is AF.
j = atan(2.09064402899/4)
j = 27.5943138509°
Where j is angle ABF.
k = 90-27.5943138509
k = 62.4056861491°
Where k is angle FBC.
l = 180-62.4056861491-atan(4/3)
l = 64.46421149674°
Where l is angle BGC.
(3/sin64.46421149674)
= (m/sin62.405686149)
m = 2.94658198738 cm.
Where m is CG.
Therefore;
Shaded Area, Triangle CFG is;
0.5*2.94658198738*4.10206390516sin24.06202137986
= 2.46410161513 cm²
