Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
Let alpha be x.
tanx = 10/a --- (1)
tan2x = a/30 --- (2)
From (2)
a = 30tan2x --- (3)
Sub. (3) in (1)
tanx = 10/30tan2x
tanxtan2x = ⅓
tanx(2tanx/(1-tan²x)) = ⅓
2tan²x/(1-tan²x) = ⅓
6tan²x = 1-tan²x
7tan²x = 1
tan²x = ⅐
x = atan(⅐√(7))°
x = 20.70481105464°
It implies;
From (3)
a = 30tan2x, and x = 20.70481105464°
a = 30tan(2*20.70481105464)
a = 26.45751311065 cm.
b = 2x
b = 41.40962210927°
c = 90-b
c = 48.59037789073°
d = 90-x
d = 69.29518894536°
e = 180-d-c
e = 62.11443316391°
(f/sin69.29518894536) = (10/sin62.11443316391)
f = 10.58300524426 cm.
g = 20-f
g = 9.41699475574 cm.
tan41.40962210927 = h/20
h = 17.63834207376 cm.
Let the yellow inscribed square side length be i.
tan41.40962210927 = i/j
j = 1.13389341903i cm.
k = i+1.13389341903i
k = 2.13389341903i
sin48.59037789073 = l/30
l = 22.5 cm.
m = (22.5-2.13389341903i) cm.
n = 17.63834207376+m
n = (40.13834207376-2.13389341903i) cm.
tan62.11443316391 = (40.13834207376-2.13389341903i)/i
i = 9.9754416633 cm.
tano = (0.5*9.41699475574)/17.63834207376
o = 14.94639616033°
k = 2.13389341903i
k = 21.28652931723
p = k-9.9754416633
p = 11.31108765393 cm.
q = 48.5903778907-x
q = 27.88556683606°
r = 45-27.88556683606
r = 17.11443316394°
s = 27.88556683606-14.94639616033
s = 12.93917067573°
t² = 17.63834207376²+9.41699475574²
t = 19.99477185018 cm.
u = 180-r-s
u = 149.94639616033°
(v/sin12.93917067573) = (19.99477185018/sin149.94639616033)
v = 8.93983473397 cm.
2w² = 8.93983473397²
w = 6.32141776308 cm.
x = 9.41699475574-w
x = 3.09557699266 cm.
tan14.94639616033 = 3.09557699266/y
y = 11.59623581162 cm.
z = 0.5*9.9754416633
z = 4.98772083165 cm.
a1 = z-3.09557699266
a1 = 1.89214383899 cm.
b1 = 40.13834207376-11.59623581162-4.98772083165-11.31108765393
b1 = 12.24329777656 cm.
It implies;
Length Red in cm, to 2 decimal places is;
√(12.24329777656²+1.89214383899²)
= 12.38864596124 cm
≈ 12.39 cm
