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Forces-Vector triangles

  • Thread starter stolencookie
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    forces vectors
  • #1
stolencookie

Homework Statement


Vector A: 1.96N at 20° Vector B: 1.71N at 65°
There are different parts I did the graph part by hand. I am having trouble finding it analytically. We have to use cm for our triangle.

Homework Equations




The Attempt at a Solution


Vector A: 1.96 cm at 20°
Vector B: 1.71 cm at 65°
Vector C: ? at 95°
180°-85°=95°
Then I had trouble after that since it isn't a right triangle. Would I use sine laws or cosine laws? I am lost.
 

Answers and Replies

  • #2
Charles Link
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If I understand the problem correctly, side ## A ## makes an angle of 20 degrees with side ## C ##, and the angle opposite side ## C ## is angle ## c= 65 ## degrees. If that is correct, the law of cosines can be used to find side ## C ##: ## \\ ## ## C^2=A^2+B^2-2AB \cos(c) ##. ## \\ ## It would help to have a diagram, but I think I got it right.
 
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  • #3
stolencookie
If I understand the problem correctly, side ## A ## makes an angle of 20 degrees with side ## C ##, and the angle opposite side ## C ## is angle ## c= 65 ## degrees. If that is correct, the law of cosines can be used to find side ## C ##: ## \\ ## ## C^2=A^2+B^2-2AB \cos(c) ##. ## \\ ## It would help to have a diagram, but I think I got it right.
Thank you, I wish i knew how to put up a triangle...
 
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