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Solving the triangle

  1. Apr 20, 2008 #1
    1. The problem statement, all variables and given/known data
    use the information to solve the triangle: b=1, c=the square root of (3)+1, B=15 degrees.
    given that sin15 degrees =the square root of 6 -the square root of 2 divided by 4. FYI.. is in the square root and than add one.

    2. Relevant equations
    The formulas given the law of sin sinA/a=sinB/b=sinC/c
    The law cosine

    3. The attempt at a solution
    I used the law of sin to find angle C. The problem is written out as the square root of (3)+1 Sin(15)/1. I came out with .70 degrees for angle c. For some reason I can't get that answer again. Also I'm not sure which formula I should be using.
  2. jcsd
  3. Apr 21, 2008 #2

    Gib Z

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    Homework Helper

    Well using the sine rule, we get [tex] \frac{\sin 15*}{1} = \frac{\sin C}{\sqrt{3}+1}[/tex].

    The * I used to denote degrees. Anyway, to get the angle C, we must get Sin C by itself on one side. Once you do that, simplify the Sin 15 degrees from what it gave you in the question, and also take out a factor of sqrt2 from it. Then it should simplify very nicely, you shouldn't even need a calculator for it. You get some nice common angle, not 0.70 degrees.
  4. Apr 21, 2008 #3


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    Staff Emeritus
    Science Advisor

    msdenise15, you forgot to take the arcsine! What you calculated was that sin(C)= .707..= sqrt(2)/2. And, as Gibz said, that's a "well known" angle.
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