Mathematics Question and Solution
Calculating the inscribed angle beta, let it be y.
Let (x+3) units be the side length of the square.
a²+x² = 4²
a = √(16-x²) units.
b = 1+a
b = (1+√(16-x²)) units.
b is the side length of the square.
Notice.
b = (x+3) units as well.
Therefore;
Calculating x.
It implies;
x+3 = 1+√(16-x²)
(x+2)² = 16-x²
x²+4x+4 = 16-x²
2x²+4x-12 = 0
x²+2x-6 = 0
(x+1)² = 6+(1)²
(x+1)² = 7
x = -1±√(7)
It implies;
x ≠ -1-√(7) units.
x = (√(7)-1) units.
x = 1.64575131106 units.
Recall.
b = x+3
b = √(7)-1+3
b = (2+√(7)) units.
b = 4.64575131106 units.
Again, b is the side length of the square.
tanc = 3/4.64575131106
c = atan(3/4.64575131106)°
c = 32.8524055273°
tand = 1/4.64575131106
d = atan(1/4.64575131106)°
d = 12.1475944727°
Therefore, the required angle beta (y) is;
y+c+d = 90
y = 90-c-d
y = 90-32.8524055273-12.1475944727
y = 45°
Again, y is the required inscribed angle beta.
