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The diagonals of rectangle PQRS intersect at point x. if PS = 6 and RS = 8, then what is sin PXS?

Mathgenius  Nov 18, 2018
 #1
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P               Q

 

6      X

 

S      8         R

 

This is similar to your other problem

 

Triangle PSR  forms a  6-8-10  Pythagorean Right triangle with PR   = 10

So....PX  =  SX  =  (1/2) (10) =  5

 

Using the Law of Cosines

PS^2  = PX^2 + SX^2  - 2(PX * SX)cosPXS

6^2 = 5^2 + 5^2  - 2 (5 * 5)cosPXS

 

36 - 2(5)^2

_________  =  cos PXS

 -2(5)^2

 

 

36  - 50

______    =   cos PXS

  -50

 

 

50 - 36                           14            7

_____   =  cos PXS  =   ___ =     ___

   50                               50           25

 

Since  PXS is acute.....then its sine will be positive  and we can find it as

 

sin PXS  = √ [ 1 - cos^2 PXS]  =  √ [ 1 -  (7/25)^2 ]  = √ [ 1 - 49/625]  =  √ [625 - 49]/ 25  =

 

√576/ 25   =    24 / 25

 

 

 

cool cool cool

CPhill  Nov 18, 2018

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