Mathematics Question and Solution
BC = √(3²+4²)
BC = 5 cm.
tana = (4/3)
a = atan(4/3)°
tan(½(atan(4/3))) = AF/3
AF = 3tan(½(atan(4/3)))
AF = 1.5 cm.
b = 90-½(atan(4/3))
b = 63.43494882292°
Where b is angle AFE.
sin63.43494882292 = AE/1.5
AE = 1.3416407865 cm.
(BE)² = 3²-1.3416407865²
BE = 2.683281573 cm.
Calculating Length CE.
(CE)² = 5²+2.683281573²-2*5*2.683281573cos(0.5atan(4/3))
CE = 2.86356421266 cm.
(2.683281573/sinc) = (2.86356421266/sin(0.5atan(4/3)))
c = 24.77514056879°
Where c is angle BCE.
Angle ECF = atan(3/4)- 24.77514056879
Angle ECF = 12.09475707705°
tan(0.5atan(4/3)) = DE/2.683281573
DE = 1.3416407865 cm.
EF = √(3²+1.5²)-2.683281573
EF = 0.67082039325 cm.
tand = 1.3416407865/0.67082039325
d = 63.43494882292°
Where d is angle EFG.
e = 180-63.43494882292-atan(3/1.5)
e = 53.13010235416°
Where e is angle CFG.
Angle CGF = 180-53.13010235416-12.09475707705
Angle CGF = 114.77514056879°
Length CF = 4-1.5
Length CF = 2.5 cm.
Calculating Length CG.
(CG/sin53.13010235416) = (2.5/sin114.77514056879)
CG = 2.20274170204 cm.
Calculating Length EG.
EG = Length CE - Length CG
EG = 2.86356421266-2.20274170204
EG = 0.66082251062 cm.
It implies;
Length CG ÷ Length EG
= 2.20274170204÷0.66082251062
= 3.33333333329
= 10/3
