AC^2 = 2^2 + 3^2 - 2(3)(2)cos(ABC)
AC^2 = 13 -12cos(ABC)
cos(ADC) =-cos(ABC)
We get...
AC^2 = 6^2 + 10^2 - 2(6)(10)cos(ADC)
AC^2 = 6^2 + 10^2+120cos(ABC)
AC^2 = 136 + 120cos(ABC) (2)
Use the subtraction method to get...
0=123+(120+12) cos(ABC)
0=123+132cos(ABC)
DB^2 = 6^2+2^2-2(6)(2)cos(DAB)
DB^2 = 40-24cos(DAB)
Therefore, DAB is obtuse
Second step!
DB^2 =10^2 + 3^2 - 2(10)(3)cos(DCB)
DB^2 =109 + 60cos(DAB)
Use the subtraction method again...
0 = 69+(60 + 24)cos(DAB)
-23/28 = cos(DAB)
We get that sin (DAB) = √ [ 1 - (23/28)^2 ] = √(255)/28 = sin PCB
(PB + AB) * PB = PC(PC+CD)
(PB + 2) * PB= 7/11)PB((7/11)PB + 10)
PB^2 + 2PB=(49/121)PB^2 + (70/11)PB
Now, let
PB = x
x^2+2x = (49/121)x^2 +(70/11)x
(121 - 49) x^2 / 121 + (2 - 70/11)x = 0
(72/121)x^2 - (48/11)x = 0
(72/11)x = 48
So the answer is 616/9
THAT IS ONE TOUGH PROBLEM!!!!