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In triangle ABC,AB = 13, BC = 14, and AC = 15. Let M be the midpoint of BC. Find AM.

 Apr 10, 2020
 #1
avatar+20981 
+1

Step 1:  Since you know the three sides of the triangle, use the Law of Cosines to find the size of angle(B).

 

Step 2:  Now, using angle(B), AB and BM, use the Law of Cosines to find AM.

 Apr 10, 2020
 #2
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i dont know what that is so could you help me out?

Guest Apr 10, 2020
 #3
 #4
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That is a totally different question.

Guest Apr 10, 2020
 #5
avatar+111329 
+2

                   A

 

        13                     15         

 

B             14                         C

 

 

BM =  7

 

Using  the Law  of  Cosines  we  have

 

AC^2  =  BC^2 + AB^2 - 2 (BC * AB) cos (ABC)

 

15^2  = 14^2  + 13^2  - 2(14 * 13) cos (ABC)

 

[ 15^2  - 14^2  - 13^2 ]  /  [ -2 (14*13) ]  =  cos (ABC)

 

-140 / -364 =  5/13  =  cos (ABC)

 

So  AM  can be found as

 

AM  =  sqrt [ BM^2  + AB^2  -  2(BM * AB) cos (ABC) ]

 

AM = sqrt  [ 7^2 + 13^2  - 2(7*13) * (5/13) ]

 

AM = sqrt [ 49 + 169  - 2 * 7 * 5 ]

 

AM  = sqrt  [ 49 +169 - 70]

 

AM = sqrt [ 148 ]  =  sqrt [ 4 * 37]  =  2 sqrt [37] units  ≈ 12.166 units

 

 

cool cool cool

 Apr 10, 2020

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