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Determining the area of a triangle

  1. Dec 8, 2015 #1

    diredragon

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    1. The problem statement, all variables and given/known data
    http://i.stack.imgur.com/KzCIl.png
    From the given picture the known quantities are:
    ##r = 7*3^{1/2} ##
    ##BC = 13 ##
    angle opposite to AC is
    ##120 ## degrees
    2. Relevant equations
    3. The attempt at a solution

    I figured i could use a sine rule to get the side ##AC ##
    ##\frac{AC}{sinb} = 2r ##
    I got ##AC = 21 ##
    I dont know where to go from this now. cosine rule to get side ##AB ## gives me weird numbers. What should be done now?
     
  2. jcsd
  3. Dec 8, 2015 #2
    Are you talking about angle B? Because if you are, that angle is clearly acute
     
  4. Dec 8, 2015 #3

    diredragon

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    Yes angle B, in the picture it is shown as acute but in a given problem its not. The picture represents the general diagram.
     
  5. Dec 8, 2015 #4

    cnh1995

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    If you get AB, you can use Heron's formula to calculate the area of the triangle.
     
  6. Dec 8, 2015 #5

    cnh1995

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    From your diagram, it is clear that the circle is circumcircle of the triangle ABC. However, since it is an obtuse angled triangle, the circumcenter should be outside the triangle. This will affect the calculations. I think you should draw and refer the exact diagram.
     
  7. Dec 8, 2015 #6

    diredragon

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  8. Dec 8, 2015 #7

    Samy_A

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    There may be a shorter method, but the sine rule can give you the angle in A.
    Then, having the three angles and two sides, finding the area should be easy.
     
  9. Dec 8, 2015 #8

    diredragon

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    What would be the easier method? I dont really want to deal with the cosine rule, there has to be simpler approach.
     
  10. Dec 8, 2015 #9

    Samy_A

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    Once you have the three angles, you don't need the cosine rule.
    Remember the basic formula for the area of a triangle? Area=base*height/2.
     
  11. Dec 8, 2015 #10

    diredragon

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    I dont see a way of getting all three angles. I have ##a ## ##b ## and an angle ##ABC ## . I tried to calculate the ##c ## side by using the cosine rule
    ##b^2 = a^2 + c^2 - 2accosB ##
    from this i get
    ##c^2 + 13c - 272 = 0 ## but the numbers dont come out right
     
  12. Dec 8, 2015 #11

    cnh1995

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    ABC is not the angle between a and b right? You can get all the three sides and angles using sine rule here.
     
    Last edited: Dec 8, 2015
  13. Dec 8, 2015 #12

    diredragon

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    I need to make a correction. I have written that ##AB = 13 ##, it actually equals ##9 ##. That then gives me fine numbers. Using the sine rule:
    ##\frac{b}{sinb}=2r ## i get ##b = 21 ##
    Then using the cosine rule
    ##b^2=a^2 + c^2 - 2accosb ## i get ##a=16 ## and ##c = 9 ## is known.
    Now the height:
    ##h_a=csin60 = \frac{9*3^{1/2}}{2} ##
    Finally ##A=\frac{(h_a)*a}{2} = 36\sqrt{3} ##
     
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