Mathematics Question and Solution
a² = 24²+7²
a = 25 units.
a is DF.
b²+20² = a²
b² = 25²-20²
b = 15 units.
b is EF.
tanc = 7/24
c = atan(7/24)°
c is angle CED.
tand = 15/20
d = atan(3/4)°
d is EDF.
e = 180-c-d
e = (180-atan(7/24)-atan(3/4))°
e = 126.8698976458°
e is angle ADE.
f = ½(180-e)
f = 26.5650511771°
f is angle DAE = angle AED.
g = 20+24
g = 44 units.
g is AC.
Therefore the required length, length AB is;
cos26.5650511771 = 44/h
h = 22√(5) units.
h = 49.193495505 units.
h is the required length, length AB.
Or
j = 7+15
j = 22 units.
j is BC.
Therefore the required length, length AB is;
√(44²+22²)
AB = √(2420)
AB = 22√(5) units.
AB = 49.193495505 units.
