+0  
 
0
49
1
avatar+55 

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

 Apr 19, 2020
 #1
avatar+20812 
+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 19, 2020

37 Online Users

avatar
avatar