Mathematics Question and Solution
Let k = 1 unit.
a = 5+1
a= 6 units.
a is the side length of the square.
b = ½(a)
b = 3 units.
c = b-1
c = 2 units.
tand = 6/2
d = atan(3)°
e = (180-atan(3))°
It implies;
(3/sin(180-atan(3))) = (1/sinf)
f = 18.4349488229°
g = 180-(180-atan(3))-f
g = 53.1301023542°
h = 90-g
h = 36.8698976458°
j² = 3²+6²-2*3*6cos36.8698976458
j = 4.0249223595 units.
(6/sink) = (4.0249223595/sin36.8698976458)
o = 63.4349488229°
l = 180-63.4349488229
l = 116.5650511771°
Therefore, the required angle, alpha is;
Let it be m.
m = 180-l-f
m = 180-116.5650511771-18.4349488229
m = 45°
Again, m is the required angle alpha.
