Mathematics Question and Solution
Notice!
The composite figure (plane shape) is not drawn to scale.
Let the side length of triangle ABC (equilateral triangle) be 1 unit.
1² = 2a²-2a²cos120
1 = 3a²
a = √(1/3)
a = ⅓√(3) units.
a is the radius of the bigger circle.
b = 2a
b = ⅔√(3) units.
b is the diameter of the bigger circle.
Let c be the radius of the smaller circle.
d = b-c
d = (⅔√(3)-c) units.
Therefore, calculating c.
sin30 = c/d
½ = c/(⅔√(3)-c)
2c = ⅔√(3)-c
6c = 2√(3)-3c
9c = 2√(3)
c = ⅑(2√(3)) units.
c = 0.3849001795 units.
Again, c is the radius of the smaller circle.
f² = e²+c²
f² = e²+(⅑(2√(3)))²
f² = e²+(12/81)
f = √(e²+(4/27)) units.
e is length AE.
g = a-c
g = ⅓√(3)-⅑(2√(3))
g = ⅑(3√(3)-2√(3))
g = ⅑√(3) units.
√(e²+(4/27))² = (⅑√(3))²+(⅓√(3))²-2*⅑√(3)*⅓√(3)cos60
e²+(4/27) = ⅓+(1/27)-⅑
e² = (9+1-3)/27-(4/27)
e² = (7/27)-(4/27)
e² = (3/27)
e = √(1/9)
e = ⅓ units.
Again, e is length AE.
tanh = c/e
tanh = 0.3849001795/(⅓)
h = atan(0.3849001795/(⅓))
h = 49.1066053538°
j = 2h
j = 2*49.1066053538
j = 98.2132107077°
j is angle BAD.
k = 180-60-j
k = 120-98.2132107077
k = 21.7867892923°
k is angle ADB.
(l/sin60) = (1/sin21.7867892923)
l = 2.3333333339 units.
l = (7/3) units.
l = 2⅓ units.
l is length AD.
Therefore,
Length AD ÷ Length AE is;
= (7/3)÷(1/3)
= (7/3)÷(3/1)
= 7
