Mathematics Question and Solution
Calculating the required angle, let it be x.
Let a be the side length of the ascribed square.
½(ab) = 1
b = 2/a --- (1).
1+1+c = ½(a²)
c = ½(a²)-2
c = ½(a²-4) square units.
d = a-b
d = a-(2/a)
d = (a²-2)/a units.
f²+d² = a²
f² = a²-((a²-2)/a)²
f² = a²-((a⁴-4a²+4)/a²)
f² = (a⁴-a⁴+4a²-4)/a²
f² = (4a²-4)/a²
f = 2√(a²-1)/a units.
g = a-f
g = a-(2√(a²-1)/a)
g = (a²-2√(a²-1))/a units.
It implies;
½(ag) = c
½*a*(a²-2√(a²-1))/a = ½(a²-4)
a²-2√(a²-1) = a²-4
2 = √(a²-1)
a² = 5
a = √(5) units.
Again, a is the side length of the ascribed square.
Recall.
b = 2/a
And a = √(5) units.
b = 2/√(5)
b = ⅕(2√(5)) units.
b is the height of the purple area inscribed triangle.
Recall Again.
g = (a²-2√(a²-1))/a
And a = √(5) units.
g = (√(5)²-2√(√(5)²-1)/√(5)
g = (5-4)/√(5)
g = ⅕√(5) units.
g is the height of the inscribed triangle with the indicated angle.
Therefore;
h = a-b
h = √(5)-⅕(2√(5))
h = ⅕(3√(5)) units.
It implies, x, the required angle is;
x = j+k
tanj = b/g
j = atan(⅕(2√(5))/(⅕√(5)))
j = atan(2)°
tank = h/g
k = atan((⅕*3√(5))/(⅕√(5)))
k = atan(3)°
Therefore;
x = atan(2)+atan(3)
x = 135°
