In triangle PQR, PQ=13 , QR=14 , and PR=18 . Let M be the midpoint of QR . Find PM.
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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
+128408 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
