Mathematics Question and Solution
Let angle BAC = 30°
Let AB = AC = 1 unit.
a² = 2-2cos30
a = 0.51763809021 units.
a is BC.
Since;
BK = 2CK, therefore;
3CK = BC
CK = 0.51763809021/3
CK = 0.17254603007 units.
It implies;
BK = 2*0.17254603007
BK = 0.34509206014 units.
b² = 1+0.34509206014²-2*0.34509206014cos(0.5(180-30))
b = 0.96977097039 units.
b is AK.
(0.96977097039/sin75) = (1/sinc)
c = 84.89609063879°
c is angle AKB.
d = 180-84.89609063879-30
d = 65.10390936121°
d is angle PBK.
(e/sin65.10390936121) = (0.34509206014/sin30)
e = 0.62604720107 units.
e is PK.
f = 180-84.89609063879
f = 95.10390936121°
f is angle AKC.
g² = 0.62604720107²+0.17254603007²-2*0.62604720107*0.17254603007cos95.10390936121
g = 0.66402333183 units.
g is CP.
Calculating angle CPK, let it be h.
(0.66402333183/sin95.10390936121) = (0.17254603007/sinh)
h = 15°
2h = 2*15 = 30°
Therefore;
2(angle CPK) = 2h = 30°
It implies;
2(angle CPK) = angle BAC.
Proved.
