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61^2=29^2+50^2-2(29)(50)cos x

Guest Mar 27, 2017
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61^2=29^2+50^2-2(29)(50)cos x   

 

Solving for cosx  we have

 

[ 61^2 - 29^2 - 50^2 ]  /  [ -2 * 29 * 50 ]    = cos x

 

Using the cosine inverse, we have

 

arccos ( [ 61^2 - 29^2 - 50^2 ]  /  [ -2 * 29 * 50 ]  ) = x   ≈  97.53°

 

 

 

cool cool cool

CPhill  Mar 27, 2017

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