tan(0.5atan(3/4))= a/24
a = 8 units.
Where a is DF.
b = 18-8
b = 10 units.
Where b is CF.
tan(0.5atan(4/3))= c/18
c = 9 units.
Where c is BE.
d = 24-9
d = 15 units.
Where d is CE.
It implies;...
(AM)² = 15²+8²-2*15*8cos60
AM = √(225+64-120)
AM = √(105+64)
AM = √(169)
AM = 13 units.
(13/sin60) = (8/sina)
a = 32.20422750397°
Where a is angle BAM.
b = 90-32.20422750397
b = 57.79577249603°
W...
Let AB = BC = CD = 1 unit.
AC = √(1²+1²)
AC = √(2) units.
(AD)² = 3-2√(2)cos105
AD = 1.93185165258 units.
Therefore;
(1.93185165258/sin105) = (√(2)/sin?)
? = 45°
Where ? is angle ADC.
a² = 4²+4²-2*4*4cos150
a = 7.72740661031 cm.
Where a is BE.
b = angle ABE = ½(180-150)
b = 15°
c = angle AFE = 180-45-15
c = 120°
(4/sin120) = (d/sin45)
d = 3.26598632371 cm.
Where d is EF.
Noti...
a = asin(1/6)
b = 2a
b = 2asin(1/6)
Where b is angle BAC
cos(2asin(1/6)) = c/6
c = ⅓(17) units.
Where c is AD.
d = 6-c
d = 6-⅓(17)
d = ⅓ units
Where d is CD.
sin(2asin(1/6)) = e/6
e = 1.97202659...
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Notice;
Since angle BDC equals angle BAC which are the opposite inscribed angles of quadrilateral ABCD.
And;
CD+BD = 40 units.
It implies;
CD+BD = AC+AB
It also implies that;
Area Triangle BD...
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9 = 8-8cosa
8cosa = -1
a = acos(-1/8)
a = 97.18075578146°
Therefore;
a• = 180-a
a• = 180-97.18075578146
a• = 82.81924421854°
b = ½(180-82.81924421854)
b = 48.59037789073°
c = ½(48.59037789073)...
