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.
a = 4.64101615138 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.
Therefore;
Length RT is;
√(a²+i²)
= √(4.64101615138²+3.39745962156²)
= 5.75167478196 cm.
