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In Triangle SPT, Q is on SP and R is on PT. We know PQ = 8, QS = 4, ST = 9, and PR = QR = 6. Find all possible values of RT. 

There was no picture. Thank you if you could help. 

 Nov 20, 2020
 #1
avatar+116125 
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We have something like this

 

                    P

          

                 8            6

                                     R        

          Q           

 

      4

S                      9                                 T

 

We  know  three sides of isisceles triangle  PQR,  PQ  = 8  and  PR and QR  = 6

 

By the Law of Cosines we can find angle SPT= angle QPR

QR^2 = PQ^2  + PR^2  - 2(PQ)(PR) cos ( SPT)

6^2  = 8^2  + 6^2  - 2(8)(6) cos (SPT)

-8^2  = - 96 cos (SPT)

64/96 =  cos (SPT)

SPT  = arccos ( 64/96)  ≈  48.19“

 

In triangle  SPT, we  know an angle (SPT)  and two sides SP  and ST

We  have SSA  (an ambiguous case).....we may have no triangle......one triangle...or  two triangles

 

Using the Law of Sines

 

sin (SPT) / ST = sin PTS / PS

sin (48.2) / 9 = sin PTS / 12

(4/3)sin 48.2 =  sin PTS

arcsin ( 4/3 * sin 48.2)  = PTS  ≈  83.7°

 

Then angle  PST  =  180  - 48.2 - 83.7  =   48.1°

And by the Law of Sines  we can find PT as

PT/ sin PST  =  ST /  sin SPT

PT / sin 48.1  = 9 / sin  48.2

PT = 9 sin 48.2/sin 48.1

PT ≈  9

So  RT  =  PT - PR =  9 - 6  = 3    

 

We may also have another triangle

Since the sine of an angle = the sine of its supplement

supplement of PTS  =  (180 - 83.7)  = 96.3°

So the remaining angle, PST  could  =  180 - 48.2 - 96.3  =  35.5°

 

So....usng the Law of Sines again

 

PT/ sin PST  = ST  / sin SPT

PT / sin 35.5  = 9/ sin 48.2

PT = 9 sin 35.5/sin48.2  ≈  7

And RT could also take the  value  PT -- PR  = 7 - 6   =  1

 

 

cool cool cool

 Nov 20, 2020

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