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In right triangle ABC, we have AB=10, BC=22, and angle ABC=90. If M is on AC such that BM is a median of ABC, then what is cos angle ABM? Express your answer in exact form.

 Apr 20, 2022
 #1
avatar+122388 
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A

 

10

 

B            22              C

 

 

If BM  is a median, then CM =   sqrt  [ 10^2 + 22^2 ]  / 2    =  sqrt (584) / 2 = 2sqrt (146) /2 =    sqrt (146)  =    AM

 

And M will be the circumcenter of the circumradius of triangle ABC....so  we have triangle ABM  where

 

AM = BM    = sqrt (146)  and AB = 10

 

Using the Law of Cosines

 

AM ^2  = BM^2 + AB^2  - 2 (BM * AB)  cos ( ABM)

 

146 = 146 + 100 =  -2 *sqrt (146)(10) cos (ABM)

 

-100 / [ -20 sqrt  (146 ) ] =   cos ABM

 

cos ABM  =   5  /sqrt (146)  =  (5/146)sqrt (146)

 

 

cool cool cool

 Apr 20, 2022

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