Mathematics Question and Solution
Calculating length PD, let it be x.
Notice.
At the inscribed triangle.
AB should be 4 cm.
While AC should be 6 cm.
Resolving with the above corrections.
5² = 6²+4²-2*6*4cosa
48cosa = 52-25
a = acos(27/48)
a = 55.7711336722°
a is angle BAC.
b = ½(a)
b = 0.5(55.7711336722)
b = 27.8855668361°
(5/sin55.7711336722) = (4/sinc)
c = 41.4096221093°
c is angle ACB.
d = 180-a-c
d = 180-55.7711336722-41.4096221093
d = 82.8192442185°
d is angle ABC.
e = 180-b-c
e = 180-27.8855668361-41.4096221093
e = 110.704811055°
e is angle APC.
f = 180-e
f = 69.2951889454°
f is angle APB.
(6/sin110.704811055) = (g/sin27.8855668361)
g = 3 cm.
g is CP.
h = CP-½(BC)
h = 3-0.5(5)
h = 3-2.5
h = 0.5 cm.
It implies, the required length PD (x) is;
cosf = x/h
cos69.2951889454 = 0.5/x
x = 0.5/(cos69.2951889454)
x = 1.41421356238 cm.
x = √(2) cm.
Again, x is the the length PD.
