Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
Let alpha be x.
Calculating Cot²x and (Red length ÷ Blue length)² as single fractions.
Let the side length of the square be 2 units.
tan30 = a/1 no
a = tan30
a = ⅓√(3) units.
b = 2-a
b = 2-⅓√(3)
b = ⅓(6-√(3)) units.
tanx = 1/b
tanx = 1/(⅓(6-√(3)))
tanx = 3/(6-√(3))°
Calculating Cot²x.
cotx = 1/tanx
cotx = 1/(3/(6-√(3)))
cotx = ⅓(6-√(3))
Therefore; cot²x is;
(⅓(6-√(3)))²
= ⅑(36-12√(3)+3)
= ⅑(39-12√(3))
= ⅓(13-4√(3))
Calculating red length. Let it be c.
c² = 1²+(⅓(6-√(3)))²
c² = 1+⅑(36-12√(3)+3)
c² = 1+⅓(13-4√(3))
c² = ⅓(16-4√(3))
c = √(⅓(16-4√(3))) units.
Calculating blue length. Let it be d.
tan30 = e/2
1/√(3) = e/2
e = ⅓(2√(3)) units.
f = 2-e
f = 2-⅓(2√(3))
f = ⅓(6-2√(3)) units.
Therefore, length d is;
cos30 = f/d
½√(3) = ⅓(6-2√(3))/d
½√(3) = (6-2√(3))/(3d)
3√(3)d = 12-4√(3)
d = (12-4√(3))/(3√(3))
d = ⅑(12√(3)-12)
d = ⅓(4√(3)-4) units.
It implies;
(Red length ÷ Blue length)² as single fractions is;
√(⅓(16-4√(3)))²÷(⅓(4√(3)-4))²
= (⅓(16-4√(3)))÷(⅑(48-32√(3)+16)
= (⅓(16-4√(3)))÷(⅑(64-32√(3))
= 3(16-4√(3))÷(64-32√(3))
= 3(4-√(3))÷(16-8√(3)))
= (3(4-√(3))(16+8√(3)))/(256-192)
= 3(64+32√(3)-16√(3)-24)/(74)
= 3(40+16√(3))/(64)
= 3(5+2√(3))/8
= ⅛(15+6√(3))
Therefore;
Cot²x = ⅓(13-4√(3))
(Red length ÷ Blue length)² = ⅛(15+6√(3))
