Second one
There may be a way better way to solve this !!!
Using the Law of Cosines extensively
-cos DCB = cos DAB
Draw DB
DB^2 = 10^2 + 3^2 - 2(10*3) cos (DCB)
DB^2 = 6^2 + 2^2 + 2(6*2) cos (DCB)
100 + 9 - 60(cosDCB) = 36 + 4 + 24cos(DCB)
69 = (24 + 60)cos (DCB)
69/84 = cos (DCB) = 23/28
sinDCB = sqrt [ 28^2 - 23^2 ] / 28 = sqrt (255)/28 = sin DAB = sin BCP
cosADC = -cosABC
Draw AC
AC^2 = 10^2 + 6^2 - 2(6 * 10)cos ADC
AC^2 = 2^2 + 3^2 + 2(2*3)cos ADC
100 + 36 - 120cosADC = 4 + 9 + 12cosADC
136 - 120cosADC = 13 + 12cos ADC
123 = 132 cos ADC
cos ADC = 123/132 = 41/44
sin ADC = sqrt (44^2 - 41^2) / 44 = sqrt (255) / 44 = sin ABC = sin PBC
sin PBC / sin BCP = PC / BP = [ sqrt (255)/ 44 ] / [ sqrt (255) / 28]
PC/BP = 28/44 = 7/11
PC = (7/11)BP
BP^2 = BC^2 + PC^2 - 2(BC * PC)cosBCP
BP^2 = 3^2 + [ (7/11)BP]^2 - 2 [ 3 * (7/11)BP ] cos DAB
BP^2 = 9 + [ (7/11) BP ]^2 - (42/11)BP (-cosDCB)
BP^2 = 9 + [ 49/121]BP^2 + (42/11)(23/28) BP
Let BP = x
x^2 - (49/121)x^2 - 69/22x - 9 = 0
(72/121)x^2 - (69/22)x - 9 = 0
Solving this for x produces x = 22 / 3 = BP