Mathematics Question and Solution
Let the radius of the ascribed semi circle be 1 unit.
a² = 2(1)²
a = √(2) units.
b = ½(a)
b = 0.5√(2) units.
b is AM = CM.
c² = 1²+1²-2*1*1cos(0.5*90)
c = 0.7653668647 units.
c is CN.
d = ½(180-45)
d = ½(135)
d = 67.5°
d is angle OCN.
e = 45+d
e = 112.5°
e is angle MCN.
f² = (0.5√(2))²+0.7653668647²-2*0.5√(2)*0.7653668647cos112.5
f = 1.2247448714 units.
f is MN.
(1.2247448714/sin112.5) = (0.7653668647/sing)
g = 35.264389682°
g is angle CMN.
h = 90-g
h = 54.735610318°
h is angle AMP.
j = 180-45-h
j = 80.264389682°
j is angle APM.
(0.5√(2)/sin80.264389682) = (k/sin45)
k = 0.5073059362 units.
k is MP.
Therefore, the required angle, alpha is;
Let it be l.
tanl = 1.2247448714/0.5073059362
l = atan(1.2247448714/0.5073059362)
l = 67.5°
Again, l is the required angle, alpha.
