1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Triangle angles

  1. Jul 17, 2014 #1
    1. The problem statement, all variables and given/known data
    Is it possible to determine all the angles in a triangle, if we only know the length of two sides?


    2. Relevant equations



    3. The attempt at a solution
    I was thinking for quite some time and I don't think it is possible. It probably is, if two sides are peprendicular but if not, I don't think so.
     
    Last edited: Jul 17, 2014
  2. jcsd
  3. Jul 17, 2014 #2

    PhysicoRaj

    User Avatar
    Gold Member

    Think about how you relate the sides and angles of a triangle?
     
  4. Jul 17, 2014 #3
    You mean the definition of the dot product?

    ##\vec{a}\cdot \vec{b}=\left \| \vec{a} \right \|\left \| \vec{b} \right \|cos\theta ##

    Would be great yeah, but I don't have the coordinates. I only have the length of the sides.
     
  5. Jul 17, 2014 #4

    PhysicoRaj

    User Avatar
    Gold Member

    A modified version of that dot product called the cosine rule comes in handy for this. Have you studied this?
     
  6. Jul 17, 2014 #5
    I have.

    ##c^2=a^2+b^2-2abcos\theta ##

    But this is a "system" of one equations with two parameters. How would you reduce the number of parameters, or better; how would you find the length of the third side?
     
  7. Jul 17, 2014 #6

    PhysicoRaj

    User Avatar
    Gold Member

    Don't you know all the sides, as in the problem?
     
  8. Jul 17, 2014 #7
    Hah. Ok, there is a mistake in the original post. I apologize.
    I only know the length of TWO sides. (I will edit my first post)
     
  9. Jul 17, 2014 #8

    PhysicoRaj

    User Avatar
    Gold Member

    You can use the law of Sines as well. I think you can eliminate the third side by using an expression for it derived from law of sines.
     
  10. Jul 17, 2014 #9
    I can eliminate the third side but than I get another angle inside the equation.

    ##\frac{a}{sin\alpha }=\frac{b}{sin\beta }=\frac{c}{sin\gamma }##
    and
    ##c^2=a^2+b^2-2abcos\gamma##

    gives me ##(a\frac{sin\gamma }{sin\alpha })^2=a^2+b^2-2abcos\gamma##
     
  11. Jul 17, 2014 #10
    Let ##b, c## be two sides of a triangle with known lengths and let ##\alpha## be the angle between them. Now consider each ##\alpha \in (0, \pi)##.
     
  12. Jul 17, 2014 #11
    Not at all possile to know the angle of triangle with two sides known.There will e infinite number of solutions .
    Just think how will you first draw the trianle with two lengths are known.First draw one line whose length is known.Then try to draw the second line starting from on edge of the first line.This second line can be drawn at any angle zero to 360 deg.So that will result in infinite number of lines .So finally finished triange will have will have infinite solutions.
     
  13. Jul 17, 2014 #12

    Char. Limit

    User Avatar
    Gold Member

    If you know the length of two sides and the angle between those two, you can figure it out. If you know the length of two sides and the angle between one of them and the third side, you can narrow it down to two possibilities. If you don't know /any/ angles, though, there's nothing you can do.
     
  14. Jul 17, 2014 #13
    Yup, I thought this may be the case yeah. :/

    Ok, thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted