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Find the length BP.

 

 Apr 28, 2022
 #1
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Draw DB  and   AC 

 

Since ABCD is a quadrilateral  inscribed in a circle  angles   DAB and DCB are supplementary

And angle  DCB is acute  so  angle  DAB is obtuse

 

Using the Law of Cosines

 

DB^2  = 3^2 + 9^2  - 2 (3)(9) cos DCB      (1)

DB^2  =  6^2  +2^2  +2 ( 6)(2) cos DCB    (2)

 

Subtract (1) from (2)

0  =  -50  +  78 cos DCB

50  = 78 cosDCB

cos DCB  =  50/78 =    25 / 39

And DCB and BCP  are supplementary

So cos BCP =  - 25 /39

arccos (-25/39)  =BCP  

 

Using the Law of Cosines again   and noting that ADC and ABC are supplementary

 

AC^2   = 6^2 + 9^2   -2(6)(9)cos ADC     (3)

AC^2  = 2^2  + 3^2  + 2(2)(3)cos ADC    (4)

 

Subtract  (3) from (4)

0  =   -104  + 120 cos ADC

cos ADC  =  104/120  =  13  /  15

arccos (13/15)= ADC =  PBC

 

 

angle BPC =  [ 180  -  angle BCP -  angle PBC]  =   [ arccos(-1)  - arccos(-25/39) -arccos (13/15)]

 

Then by the Law of Sines

 

BP  / sin [ BCP]  = 3 / sin [ BPC]

 

BP  =  sin (arccos (-25/39) ) * 3   / sin [ arccos(-1) - arccos (-25/39)  -arccos (13/15) ]  = 20 / 3

 

 

cool cool cool

 Apr 28, 2022

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