In cyclic quadrilateral ABCD, AB = 2, BC = 3, CD = 10, and DA = 6. Let P be the intersection of lines AB and CD. Find the length BP.
+2094 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!!!!
+500 Not sure if this is a coincidence, but CTG's explanation goes exactly in tandem with Cphill's previous explanation of this problem. If you want to see for yourself, the original link is here:
https://web2.0calc.com/questions/pls-help-due-tmrow_10
Now that I take a closer look, it seems that CTG's explanation is the exact same as Cphills. @CalTheGreat if you're going to use Cphill as a reference, I'd recommend linking his explanation first and foremost! Again, this is just a big "if"
+2094 Honestly... this is one of the only ways to solve the problem....... it might be similar...
+500 I know, I'm just saying "if". However, the equations are formatted the exact same way(the amount of spacing is the exact same which caught my attention), with the logic and steps being the same? I'm not too sure.....
+2094 Really??? I would NEVER copy anyone, though. I know how serious plagiarism is. My equations are formatted differently with different steps...Chris has way more equations than me!
+500 Side by side comparison?
https://web2.0calc.com/api/ssl-img-proxy?src=%2Fimg%2Fupload%2Fbc435091ccbca0f0af8a1387%2Fcphillanswer.png&l=1
you changed one thing, and that's the brackets in this part alone. I could compare other parts, but this is just a petty argument in general. If you know you copied someone else's work, that's ok, just make sure to credit them. That's my point here. End of discussion.
+2094 If you notice, however, the equation was written slightly differently, but I see what you mean. Sorry if you thought so. As you say, end of discussion.
+118608 LOL
Cal you were sprung!
It is good that jfan caught you out. Everything he said is correct.
You must give credit where credit is due.
However, I do believe that you are using this site to learn.
And I do believe that you understood and learned from CPhill's answer.
I know CPhill will be very pleased that you are learning from his answers.
