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thank you so much for helping!! Please quick it is due :)

 May 21, 2019
 #1
avatar+17774 
+2

Using the Law of Cosines for both triangle(ABC) and triangle(ABM):

 

42  =  32 + (2a)2 - 2(3)(2a)cos(B)                             (2a)2  =  a2 + 32 - 2(3)(a)cos(B)

16  =  9  + 4a2 - 12a·cos(B)                                         4a2  =  a2 + 9 - 6a·cos(B)

7 - 4a2  =  -12a·cos(B)                                                 3a2 - 9  =  -6a·cos(B)

cos(B)  =  (4a2 - 7) / (12a)                                                         cos(B)  =  (9 - 3a2) / (6a)

 

Combining:  (4a2 - 7) / (12a)  =  (9 - 3a2) / (6a)

                       4a2 - 7  =  18 - 6a2

                            10a2  =  25

                                 a  =  sqrt(2.5)

                              BC  =  2·sqrt(2.5)

 May 21, 2019
 #2
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0

Hectictar: Can the "Apollonius Theorem" be used to solve this triangle?

 May 22, 2019
 #3
avatar+8209 
+2

I have never heard of Appollonius's Theorem, but does seem like it can be used here.

 

Using the information from here:  https://en.wikipedia.org/wiki/Apollonius%27s_theorem

 

32 + 42  =  2( (2a)2 + a2 )  
25  =  2( 4a2 + a2 )

 

 

25  =  2( 5a2 )  
25  =  10a2

 

 

2.5  =  a2  
a  =  √[ 2.5 ]

 

 

BC  =  2a  =  2√[ 2.5 ]  
hectictar  May 22, 2019
 #4
avatar+101420 
+2

I might have used Geno's approach, too.....but  the theorem used by hectictar is certainly handy!!!!

 

 

cool cool cool

 May 22, 2019
 #5
avatar+8209 
+2

The only way I could think of to do it was with the Law of Cosines too, but I will definitely try to remember Appollonius's Theorem from now on laugh

hectictar  May 22, 2019

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