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In triangle PQR, PQ=13 , QR=14  , and PR=18 . Let M be the midpoint of QR . Find PM.

 Jun 23, 2021
 #1
avatar+320 
+4

Hey there, Guest!

 

It appears that the question that you asked has already been answered here: https://web2.0calc.com/questions/more-hw-please-help.

 

In case if you're one of the people online of which don't like links, this is what it says:

 

In triangle PQR, PQ=13, QR=14, and PR=15. Let M be the midpoint of QR. Find PM.

 

Let angle PRQ be  >β<

 

1)   Let's find β  first:             cos(β) = (15² + 14² - 13²) / 2*15*14       β = 53.13°

 

2)   Now we can find  PM      (PM)² = 7² + 15² - 2*7*15 * cos(β)    

 

                                                 PM = 12.166     

Answerer: Dragan

 

Hope this helped! :)

( ゚д゚)つ Bye

 Jun 23, 2021
 #2
avatar+121000 
+1

                   P

           13                     18

 

Q                      M           7              R

 

Using the Law of Cosines twice

 

(PQ^2 - QR^2  - PR^2)  /  (-2 QR * PR)  =  cos  (PRQ)

 

( 13^2  - 14^2  - 18^2)  / ( - 2 * 14 * 18)  =   cos (PRQ)

 

              39  / 56   =  cos (PRQ)

 

And

 

PM^2  =   MR^2  + PR^2   - 2 ( MR * PR)  cos (PRQ)

 

PM^2  =  7^2  +  18^2  - 2 ( 7 * 18)   (39/56)

 

PM^2  =  197.5

 

PM  = sqrt  (202)   ≈   14.05

 

 

cool cool cool

 Jun 23, 2021

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