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Trigonometry

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

Oct 18, 2017

#1
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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"

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
+95883
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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

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
+95883
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Your rearrangement is correct if  side "c"  is opposite "theta"

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

Oct 18, 2017