+0  
 
+1
29
1
avatar+491 

In right triangle $ABC$, we have $AB = 10$, $BC = 24$, and $\angle ABC = 90^\circ$. If $M$ is on $\overline{AC}$ such that $\overline{BM}$ is a median of $\triangle ABC$, then what is $\cos \angle ABM$?

 Apr 20, 2021
 #1
avatar+118505 
+1

AC = sqrt ( 24^2  + 10^2)  = sqrt [  576 + 100 ] = sqrt (676)  = 26

So....  AM  =  13

 

And  a  median drawn  from  the hypotenuse to the opposite side  has (1/2) length of  the  hypotenuse  

So....BM =  13

 

By  the Law of Cosines

 

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

 

13^2  = 10^2  + 13^2  - 2 ( 10 *13) cos ABM

 

169  = 100  + 169  -  260 cos ABM

 

-100 / -260  = cos ABM

 

5 / 13   = cos ABM

 

cool cool cool

 Apr 20, 2021

16 Online Users