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Help with law of sines/solving triangles

  1. Dec 10, 2014 #1
    1. The problem statement, all variables and given/known data
    L2ZLKEM.png

    2. Relevant equations
    SinA/a = SinB/b = SinC/c

    3. The attempt at a solution
    a=38, c=44, ∠A = 35°
    What I have so far:
    sin(35)/38 = sinC/44
    44sin(35) = 38sinC
    44sin(35)/38 = sinC
    sinC = 0.6641
    sin^-1(0.6641)
    ∠C = 41.6º
    Is this ∠C1?
    After this I don't know how to derivive the rest.
    ---------------------------------------------
    For the second problem:
    a = 76, b = 105, ∠A = 29°
    sin(29)/76 = sinB/105
    105sin(29) = 76sinB
    sinB = 105sin(29)/76
    sin^-1(0.6698)
    ∠B = 42.1º
    Again, after this point I am lost
    I could really use some instruction on how to solve.
    Thank you
     
  2. jcsd
  3. Dec 10, 2014 #2

    jedishrfu

    Staff: Mentor

    If you know two angles of a triangle what do you know about the third angle?
     
  4. Dec 10, 2014 #3
    Oh of course! Euclidean triangles equal 180º therefore for the first one 180-35-41.6 = 103.6º
    and the second is 180-29-42.1 = 108.9
    What must I figure out after this though?
     
  5. Dec 10, 2014 #4

    jedishrfu

    Staff: Mentor

    Your choice of which angle is B is arbitrary right? So what would you choose as B1 so that B1 > B2 ?
     
  6. Dec 10, 2014 #5
    If you mean where the angle is placed on the triangle, yeah I think it's arbitrary. If angle B1 is 103.6º and B2 must be larger than I suppose 105º? not sure if I am understanding the question here.
     
  7. Dec 10, 2014 #6

    jedishrfu

    Staff: Mentor

    You only have one triangle colum one labels the angles one way and column two labels a different way.

    Consider a 30-60-90 triangle and I say A is 90 then B is 30 or its 60 right and which ever you choose then C is the other angle. Your problem is constrained by the sin rule with A and a defined.
     
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