Mathematics Question and Solution
Let the side length of the ascribed square be 1 unit.
Therefore, the side length of the inscribed green regular triangle 1 unit.
a = 90-60°
a = 30°
b² = 2-2cos30
b = 0.51763809021 units.
c = ½(180-30)
c = 75°
d = c+60
d = 135°
(1/sin135) = (0.51763809021/sine)
e = 21.4707014327°
f = 180-135-e
f = 45-21.4707014327
f = 23.5292985673°
(1/sin135) = (g/sin23.5292985673)
g = 0.56457945531 units.
h = 1-g
h = 1-0.56457945531
h = 0.43542054469 units.
j² = 0.43542054469²+1²-2*0.43542054469*1cos30
j = 0.65986403499 units.
j is the side length of the inscribed red regular triangle.
(0.65986403499/sin30) = (1/sink)
k = 49.2646492843°
l = 60-k
l = 60-49.2646492843
l = 10.7353507157°
Therefore;
m² = 0.65986403499²+0.56457945531²-2* 0.65986403499*0.56457945531cos10.7353507157
m = 0.14872692587 units.
It implies, the required angle alpha is;
Let it be n.
(0.14872692587/sin10.7353507157) = (0.56457945531/sinn)
n = 45°
