Mathematics Question and Solution
Let a be the radius of the inscribed circle.
Calculating a.
tan30 = a/6
a = 2√(3) cm.
b = inscribed circle diameter.
b = 4√(3) cm.
12² = c²+6²
c = 144-36
c = √(108)
c = 6√(3) cm.
d = 6√(3)-4√(3)
d = 2√(3) cm.
Where d is AD.
e² = 12²+(2√(3))²-2*12*2√(3)cos30
e = 2√(21) cm.
Where e is BD.
(2√(21)/sin30) = (12/sinf)
f = 139.10660535087°
Where f is angle BDA.
g = 180-30-139.10660535087
g = 10.89339464913°
Where g is angle ABD.
h = 60-10.89339464913
h = 49.10660535087°
Where h is angle FBC.
i = 90-49.10660535087
i = 40.89339464913°
Where i is angle EDO.
cos40.89339464913 = j/(2√(3))
j = 2.61861468283 cm.
k = 2j
k = 5.23722936566 cm.
Where k is DE.
l = 2√(21)-5.23722936566
l = 3.92792202425 cm.
Where l is BE.
m² = 3.92792202425²+12²-24*3.92792202425cos49.10660535087
m = 9.88505365257 cm.
Where m is CE.
(9.88505365257/sin49.10660535087) = (3.92792202425/sinn)
n = 17.48017020278°
Where n is angle BCE.
o = 180-17.48017020278-49.10660535087
o = 113.41322444635°
Where o is angle BEC
p = 180-113.41322444635
p = 66.58677555365°
Where p is angle DEF.
q = 66.58677555365-40.89339464913
q = 25.69338090452°
Where q is angle OFE = angle CFG.
cos25.69338090452 = r/(2√(3))
r = 3.12159589029 cm.
s = 2r
s = 6.24319178057 cm.
Where s is EF.
t = 9.88505365257-6.24319178057
t = 3.641861872 cm.
Where t is CF.
u = 180-25.69338090452-17.48017020278
u = 136.8264488927°
Where u is angle CGF.
It implies;
Calculating Length FG.
Let it be v.
(v/sin17.48017020278) = (3.641861872/sin136.8264488927)
v = (3.641861872sin17.48017020278)÷sin136.8264488927
v = 1.59881613006 cm.
And;
v = FG
Therefore;
FG = 1.59881613006 cm.
