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How would I write the cosine law if I was trying to find theta given all sides?

Guest Oct 18, 2017
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5+0 Answers

 #1
avatar+78618 
+2

Let  the sides be  a, b and c

 

So we have that

 

a^2 =  b^2 + c^2 - 2bc *cos (theta)     rearrange as

 

2bc * cos(theta)  =  b^2 + c^2 - a^2      divide both sides by   2bc

 

cos (theta)  =  [  b^2 + c^2 - a^2 ] /  [2bc ]

 

Take the cosine inverse

 

cos-1 (  [  b^2 + c^2 - a^2 ] /  [2bc ] )  =  theta

 

Remember that "theta"  is the angle opposite  side "a"

 

 

cool cool cool

CPhill  Oct 18, 2017
edited by CPhill  Oct 18, 2017
 #2
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Wouldn't C be subtracted by B and A and then divided by 2AB? Or is my new formula, after some deriving and algebra I did in the time it took to answer this question (Very impressed by your quick response! Thank you!!!) wrong?

C^2 - B^2 - A^2 / 2AB = Theta

Guest Oct 18, 2017
 #3
avatar+78618 
0

Notice from this point 

 

a^2 =  b^2 + c^2 - 2bc *cos (theta)  

 

I subtracted a^2  from both sides and added 2bc *cos (theta)   to both sides giving us

 

2bc * cos(theta)  =  b^2 + c^2 - a^2 

 

 

cool cool cool

CPhill  Oct 18, 2017
 #4
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I am so sorry, but I still don't understand. Would you please mind elaborating on what I did wrong?

Guest Oct 18, 2017
 #5
avatar+78618 
0

Your rearrangement is correct if  side "c"  is opposite "theta"

 

In my answer, I arbitrarily made  "a"  the opposite side to "theta"

 

 

cool cool cool

CPhill  Oct 18, 2017

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