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

 

Thanks!!!

 Apr 20, 2020
 #1
avatar+20810 
+1

In triangle(ABC), AB = 13, BC = 14, and AC = 15.   

M is the midpoint of BC.

Find AM.

 

I'm going to use the Law of Cosines twice  

-- first, with triangle(ABC), to find angle(C)

-- and then, with triangle(ABM), to find AM.

 

Triangle(ABC):  AC2  =  AB2 + BC2 - 2·AB·BC·cos(B)

--->                     152  =  132 + 142 - 2·13·14·cos(B)

--->                     225  =  169 + 196 - 364·cos(B)

--->                    -140  =  -364·cos(B)

--->                 cos(B)  =  0.3846  

--->                        B  =  67.38o  

 

Triangle(ABM)  AM2  =  AB2 + BM2 - 2·AB·BC·cos(B)

--->                   AM2  =  132 + 72 - 2·13·7·cos(67.38o)

--->                   AM2  =  148

--->                   AM  =  sqrt(148)

 Apr 20, 2020
 #2
avatar+127 
+1

Nice Job @geno3141! Can you help with my problem, it is just above this one.Thanks!

CPhilFanboy  Apr 20, 2020

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