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In triangle PQR, the side lengths are PQ=9, PR=10, and QR=17. Let X be the intersection of the angle bisector of angle P with side QR, and let Y be the foot of the perpendicular from X to side PR. Compute the length of XY.

 Feb 20, 2020
 #1
avatar+109740 
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See  the   following  :

 

 

 

 

 

 

Since  PX  is an angle bisectior....we have  the following relationnship  :

 

PR  / PQ  = XR /XQ    so

 

10/9  = XR / XQ

 

And  XR  + XQ  =   QR   =17

 

So

 

XR  = (10/19)*17  =   171/19    =  9

 

And using the Law of Cosines

 

9^2   = 17^2  + 10^2  - 2(17*10)*cos (PRQ)

 

81  - 289  - 100  =  -340*cos *PRQ)

 

-308 / -340  = cos (PRQ)   = 77/85

 

So....  sin (PRQ)  =  sqrt ( 1 - (cos (PRQ)^2 )   =  sqrt ( 1  - (77/85)^2 )   =    36/85

 

And by the Law of Sines

 

XR /sin (90)  =  XY / sin (PRQ)

 

9 / 1  =  XY / (36/85)

 

9 (36/85)  =  XY   =   324 / 85   ≈  3.81

 

 

cool cool cool 

 Feb 20, 2020
 #2
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0

That isn't correct.

 Feb 20, 2020

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