Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Operations Involving Law of Sines and multiple Unknown Values

  1. Sep 21, 2010 #1
    Hello, I was working on a computer graphics problem when I encountered an interesting sceanrio:

    I have two vectors a and b, such that the angle between them is 45 degrees.

    The vector a+b and a have an angle between them that is 30.

    This produces a problem in 'drawing' a triangle, when trying to solve this problem using the law of sines because the 'triangle' involving vectors a, b, and a+b do not 'add up' to 180 degress because we already have two angles (45 and 30) totalling 75 degrees, thereby needing an angle > 90 within this particular triangle for a sum total of 180 degrees for the triangle (since an individual angle of a triangle cant be more than 90 degrees.

    In this case, if the length of |a| = 6
    then I have one unknown angle (between b and (a+b) and two unknown magnitudes (|b| and |a+b|)

    To therefore 'build' a triangle to solve this problem using the Law of Sines, can I simply 'divide' the angle between a and b or a and a+b to create my triangle, to find |b|?

    I appreciate any insight of approaching this problem from the angle of using the Law of Sines (no pun intended), and I hope I discribed my problem clearly enough
     
  2. jcsd
  3. Sep 21, 2010 #2

    mathman

    User Avatar
    Science Advisor

    Let a, b, c be the sides and A, B, C be the opposite angles. c=a+b. I will use | | for side length.

    Your description has C=135 (180-45) and B=30. Therefore A=15. Now use the law of sines to get |b| and |c|.
     
  4. Sep 21, 2010 #3
    Mathman,

    Sigh, thanks...I looked over this problem and didnt see to approach the problem in that simple and elegant manner...thank you...but how did you know to find C= 135 where (180 -45) = 135?...its probably a simple operation too...thanks again
     
  5. Sep 22, 2010 #4

    HallsofIvy

    User Avatar
    Science Advisor

    Yes, that's exactly right. Draw a diagram of the vector addition and you will find that the "inside" angle of the triangle is the supplement of the angle between the vectors.
     
  6. Sep 22, 2010 #5
    Thanks all you guys...you are correct---and I did that last night. I did do that exact same operation...didnt realize the term was called 'supplement' (I suppose by the paralleogram law?)

    In addition, I researched this problem, or type of problems, and is this considered one of the 'Ambigous Triangle Case' types?

    I had never heard of 'Ambiguous Case' in relation to triangles, so it was an interesting read.
     
  7. Sep 22, 2010 #6

    mathman

    User Avatar
    Science Advisor

    This is not "Ambiguous case". If you remember the congruent theorems in elementary geometry, they all are based on equality involving three corresponding parts between the two triangles, at least one of which is a side. These are usually described by initials (S for side and A for angle), namely SSS, SAS, ASA, and AAS. Note that SSA is not included, since there are, in general, two different triangles with equality for two sides and an angle (acute) other than the included angle. This is the "ambiguous case".
     
  8. Sep 24, 2010 #7
    Hello,

    Would any of you be able to verify my 'proof' of a similar situation?

    Here it goes:

    If the angle between the vectors a and b is 60 degrees and |a| = 6
    and the angle between a-b and a is 30 degrees --->

    then the |b| can be found as follows:

    [1] If vectors a, b, and a-b form a Triangle by vector addition such that the angles of
    a + b + (a-c) = 180

    [2] then it follows that the angle between b and a-b is 90 degrees

    Therefore if |a| = 6 and by using the law of sines we can find |b| by

    sin (90)/6 = sin (30)/|b| ....such that sin(A)/|a| = sin(B)/|b| and statement [2]

    thus, |b| = 3

    Is this a reasonable solution or are any of my assumptions / statements misguided?
    In addition, is there any reason why one would need to use the Law of Cosines to figure this out? From a program perspective, this appears to be a much more efficient method (Law of Sines I mean)
    Thanks for any help.
     
  9. Sep 24, 2010 #8

    mathman

    User Avatar
    Science Advisor

    Slight error - I presume the 'c' should be 'b'.

    It looks OK otherwise.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook