+0  
 
0
50
1
avatar

In right triangle ABC, we have AB=10, BC=24, 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.

 Feb 27, 2020
 #1
avatar+109486 
+1

AC   = sqrt  (24^2 + 10^2)  =   sqrt  (676)  =  26

 

Let  B   =(0,0)

Let A  =(0,10)

Let C  = (24,0)

 

M  = [ (24 + 0)/2 , (10 + 0) /2 ]  =   ( 12,5)

 

BM  =  sqrt [ ( 12 - 0)^2  + ( 5 - 0)^2  ]  =  sqrt ( 144 + 25)  =  sqrt (169)   = 13

 

sin  BAC  = sin BAM   =  24 / 26  =  12/13

 

Using  the Law  of Sines

 

sin BAM  / BM  = sin ABM / AM

 

(12/13) / 13  = sin ABM / 13

 

sin (ABM)  = 12/13

 

So

 

cos ABM  =  sqrt  ( 1  - (12/13)^2 )  =  sqrt  ( 1 - 144/169)  =  sqrt  (169 -144) / 13  = 

 

sqrt ( 25) /13   =  

 

5 /13

 

 

 

cool cool cool

 Feb 27, 2020

40 Online Users

avatar
avatar
avatar
avatar
avatar
avatar
avatar