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Creating specified triangle

  1. May 5, 2006 #1
    how to create such a triangle in geometric way:

    [AB]=5.5 cm, [AC]=3.7cm, ABC=100 (angle)

    thanks
     
  2. jcsd
  3. May 5, 2006 #2

    berkeman

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    Not sure I understand the question, but check out the Law of Sines and the Law of Cosines. One of those might work for you....
     
  4. May 6, 2006 #3
    no , what it's required is that to create with geometric way so not using trigonometric laws, i need to get that triangle by using only ruler and may be circuler
     
  5. May 8, 2006 #4

    berkeman

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    I'm not sure what a circuler is, but if you can use a ruler, then you just have to think about how to construct the 100 degree angle using the ruler. Why would knowing the arctan(10 degrees) help in this?
     
  6. May 8, 2006 #5

    Gokul43201

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    ee : You must tell us what you have tried and where you are stuck. Please read the guidelines for posting here.
     
  7. May 8, 2006 #6
    Can you use both a protracter and a compass (or a compass that measures angles)?
     
  8. May 9, 2006 #7
    yes,I can use them , you said both protractor and compass even though protractor itself is used to measure angles .... anyway to where are you going to get me?
     
  9. May 9, 2006 #8

    Gokul43201

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  10. May 9, 2006 #9

    VietDao29

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    Ok, I think I'll give you some hints to start off your problem:
    First, given that AB = 5.5 cm
    Can you construct the line segment AB?
    From there, you know that: [tex]\widehat{ABC} = 100 ^ \circ[/tex]. Can you find two rays, of which every point makes with AB an angle of 100o? You also know that AC = 3.7 cm. How can you find C?
    Can you go from here? :)
     
  11. May 9, 2006 #10
    yes i know that , i tried to follow this way before you tell me about it, the problem is how to find C because [AC]=3.7cm and the angle more than 90 (100 c) makes that length of 3.7cm not achievable especially that [AC]<[AB].....that's why i'm asking here , of course i don't want to get the others solve my problem , i'm just asking for guidance.

    many thanks for you all
     
  12. May 9, 2006 #11

    berkeman

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    Well, I just sketched the triangle, and BC is about 7-something long. The triangle looks fine. So the problem is not that such a triangle doesn't exist, right EE? The problem is how to construct it with a ruler, protractor and compass, right?

    So draw line AB from the origin to the right 5.5cm long. Then mark a point 3.7cm to the right of the origin and on the segment AB. Strike an arc with your compass so that it makes a circle of radius 3.7cm centered on the origin. So now point C has to lie on that line somewhere (actually, two somewheres), right? Now your only problem is how to construct a 100 degree angle. That's why I asked early on about the arctan(10 degrees)....
     
  13. May 9, 2006 #12
    Am I wrong to suggest that the question should have stated BC is 3.7 as opposed to AC is 3.7?...as VietDao suggests, draw two rays from B where the angle it makes with AB is 100 deg and you're gonna have difficulties making AC
     
    Last edited: May 9, 2006
  14. May 9, 2006 #13

    berkeman

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    Don't you draw two rays from A? Although I guess I see what you mean, if "angle ABC" is supposted to have B as the common point of AB and BC. EE, are you sure it said "angle ABC = 100 degrees"? Or could it have meant angle CAB = 100 degrees?
     
  15. May 10, 2006 #14

    Gokul43201

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    This is all you should have said in the beginning.

    You are perfectly correct. The triangle with the given dimensions does not exist on a plane. There is an error in the question. Take it back to the teacher and ask them to correct it.
     
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