Mathematics Question and Solution
Let the inscribed square side be a.
tan60 = a/b
a = √(3)b cm.
Where b is BS = CR
It implies;
b+√(3)b+b = 10
(2+√(3))b = 10
b = 10(2-√(3)) cm.
Therefore;
a = √(3)*10(2-√(3))
a = 10(2√(3)-3) cm.
sin60 = 10(2√(3)-3)/c
c = 20(2-√(3)) cm.
Where c is CQ.
d² = 10²+(10(2-√(3)))²-20*10(2-√(3))cos60
d = 8.96575472168 cm.
Where d is AS.
(8.96575472168/sin60) = (10(2-√(3))/sine)
e = 15°
Where e is angle PAS.
f = 180-60-15
f = 105°
Where f is angle BSA.
g = 105-90
g = 15°
Where g is angle PSA.
tan15 = h/10(2√(3)-3)
h = 1.24355652982 cm.
i = 10(2√(3)-3)-1.24355652982
i = 3.39745962156 cm.
j = 60-15
j = 45°
Where j is angle SAQ.
(3.39745962156/sin45) = (k/sin60)
k = 4.16102124727 cm.
l² = 4.16102124727²+10²-20*4.16102124727cos45
l = 7.64646133895 cm.
(4.16102124727/sinm) = (7.64646133895/sin45)
m = 22.63074021249°
n = 60-22.63074021249
n = 37.36925978751°
tan37.36925978751 = o/10(2-√(3))
o = 2.04634925988 cm.
p = 10(2√(3)-3)-2.04634925988
p = 2.5946668915 cm.
Therefore;
Area S1 ÷ Area S2 is;
(0.5*2.5946668915*3.39745962156) ÷ (0.5*10(2-√(3))*2.04634925988)
= 4.40763799763÷2.74158815808
= 1.60769515459
≈ 1.61 to 2 decimal places.
