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Showing area of triangle

  1. Sep 25, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that for all [tex]\theta \epsilon (0, \pi)[/tex], the area of a triangle with side lengths a and b with included angle [tex]\theta is A = \frac{1}{2} a b sin \theta[/tex]. (Hint: You need to consider two cases)

    2. Relevant equations



    3. The attempt at a solution

    I have just begun working on this problem.. not really sure where to start.

    Does [tex]\theta \epsilon (0, \pi)[/tex] mean that the angle is > than 0 and < than pi?
    Am I supposed to show that when the angle is less than or greater than the condition then the equation to find area is not valid?
     
    Last edited: Sep 25, 2009
  2. jcsd
  3. Sep 25, 2009 #2

    Dick

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    Yes, that's what (0,pi) means. The only cases where the area is not ab*sin(theta) is where sin(theta) might be negative. They aren't in (0,pi). What's the area in that case?
     
  4. Sep 27, 2009 #3
    The area is bh/2
     
  5. Sep 27, 2009 #4

    Dick

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    They want you to give an answer in terms of the sides a and b. Not the base and the height.
     
  6. Sep 27, 2009 #5
    Can you give me a little more hint -_-;

    What are the two cases that I need to consider?
     
  7. Sep 27, 2009 #6

    Dick

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    Use trig and A=bh/2. What's h in terms of a and the included angle? Draw a right triangle. And I'm really not sure what the 'two cases' they are talking about are.
     
  8. Sep 28, 2009 #7
    h = b(sin theta)
    or
    h = b(sin 180 - theta)
     
  9. Sep 28, 2009 #8

    Dick

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    sin(theta) and sin(180-theta) are the same number. Aren't they?
     
  10. Sep 28, 2009 #9
    So can I show this by drawing a picture?
     
  11. Sep 28, 2009 #10

    Dick

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    There's a variety of ways to draw a picture to show sin(pi-x)=sin(x). Which sort did you have in mind? How do you picture sin(x)?
     
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