Mathematics Question and Solution
Calculating Alpha.
Let it be x.
Notice.
BCDE and ABFG are both square.
Let the two equal red lengths be √(2) units each.
a²+a² = √(2)²
2a² = 2
a² = 1
a = 1 unit.
a is the side length of square BCDE.
b = y+a
b = (y+1) units.
b is length AC.
y is the side length of square ABFG.
Calculating y.
√(2)² = y²+(y+1)²
2 = y²+y²+2y+1
2y²+2y-1 = 0
y²+y = ½
(y+½)² = ½+(½)²
(y+½)² = ¾
y = -½±√(¾)
y = -½±½√(3)
It implies;
y ≠ -½(1+√(3)) units.
y = ½(√(3)-1) units.
y = 0.36602540378 units.
Again, y is the side length of square ABFG.
Recall.
b = (y+1) units.
And y = 0.36602540378 units.
b = 0.36602540378+1
b = 1.36602540378 units.
tanc = (0.36602540378/
1.36602540378)
c = atan(0.36602540378/
1.36602540378)
c = 15°
Therefore x, angle alpha is;
90-45-c
x = 45-15
x = 30°
