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State one form of the Law of Cosines and provide a trick for writing the other two forms and explain when Law of Cosines should be used to solve a triangle.

User Foxichu
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Solving for Angles


\displaystyle (a^2 + b^2 - c^2)/(2ab) = cos∠C \\ (a^2 - b^2 + c^2)/(2ac) = cos∠B \\ (-a^2 + b^2 + c^2)/(2bc) = cos∠A

* Do not forget to use the inverse function towards the end, or elce you will throw your answer off!

Solving for Edges


\displaystyle b^2 + a^2 - 2ba\:cos∠C = c^2 \\ c^2 + a^2 - 2ca\:cos∠B = b^2 \\ c^2 + b^2 - 2cb\:cos∠A = a^2

You would use this law under two conditions:

  • One angle and two edges defined, while trying to solve for the third edge
  • ALL three edges defined

* Just make sure to use the inverse function towards the end, or elce you will throw your answer off!

_____________________________________________

Now, JUST IN CASE, you would use the Law of Sines under three conditions:

  • Two angles and one edge defined, while trying to solve for the second edge
  • One angle and two edges defined, while trying to solve for the second angle
  • ALL three angles defined [of which does not occur very often, but it all refers back to the first bullet]

* I HIGHLY suggest you keep note of all of this significant information. You will need it going into the future.

I am delighted to assist you at any time.

User Thomas John
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