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Draw a triangle

  1. Aug 18, 2014 #1
    1. The problem statement, all variables and given/known data
    Side ##a=4 cm##, altitude to side a is 3 cm , angle ##\alpha =60 °##.

    How can I draw that? Step by step


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 18, 2014 #2

    Mentallic

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    What are you allowed to use in your construction? A ruler and protractor for example would make this a trivial task.

    Also, have you tried anything?
     
  4. Aug 18, 2014 #3
    Am... Trivial?

    :D

    I tried:
    Decided where my point A is. :D
    Than using a thumbscrew compass I measured the altitude to side a and drew a circle around point A.
    I just picked a random radius of that circle, and drew a perpendicular line to it. (on this line, my a side should be).
    However, the problem with this is I can't find a way to exactly determine where the other two points are now...
     
  5. Aug 18, 2014 #4

    Mentallic

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    Ok hang on, just so I have a clear understanding of what your triangle dimensions are.

    If we draw side a=4cm alone the bottom, let A be the left point, B the right point (hence AB is 4cm), and C the top point that will connect the triangle together. Now which length is 3cm long and which angle is 60o?
     
  6. Aug 18, 2014 #5
  7. Aug 18, 2014 #6
    If angle ## \alpha ## is adjacent to side a it is easy, but I assume this is not the case.

    So you have two right angled triangles one with an angle of ## \theta ## adjacent to a side of length 3 cm and opposite a side of length ## x ##, and one with an angle of ## 60 - \theta ## adjacent to a side of length 3 cm and opposite a side of length ## 4 - x ## cm.

    Two equations, two unknowns - and a very surprising (to me) solution.

    Not sure if you can do this with ruler, protractor and compass though.
     
  8. Aug 18, 2014 #7

    Mentallic

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    You'll be surprised with how much time is spent wasted because the problem was misunderstood and the standards of triangle labelling aren't always upheld.

    You can use the sine and cosine rule to find length c. If b = 3cm then with the cosine rule

    [tex]c^2 = a^2+b^2-2ab\cdot \cos C[/tex]

    Plugging our known values in

    [tex]c^2=3^2+4^2-2\cdot 3\cdot 4\cos C[/tex]

    [tex]c^2=25-24\cos C[/tex]

    Now for the sine rule

    [tex]\frac{\sin A}{a} = \frac{\sin B}{b}[/tex]

    Plugging values

    [tex]\sin B = \frac{3\sqrt{3}}{8}[/tex]

    Hence you've got B and then you can find [itex]\sin C[/itex]. I'm assuming you're only using a compass, because if you could use a protractor then just the sine rule would be sufficient to solve your problem.
     
  9. Aug 18, 2014 #8
    Not really. :D
    I understood it exactly the same way you did. But I later found out that it is not so simple. I am sorry for being evil.
    That works perfectly!
    Thank you!
     
  10. Aug 18, 2014 #9
    Ok, or maybe it doesn't work perfectly.

    Who said b=3cm?
    If you used b just for labelling, than I am sorry to say that in that case the value of a is no longer 4 cm.

    Or is there something I missunderstood?
     
  11. Aug 18, 2014 #10

    Mentallic

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    EDIT: incorrect post ahead.


    Ugh... altitude = 3cm... I took that as being the side length of the triangle. Sorry, it's been a long night. That's another thing that'll waste everyone's time :biggrin:

    Just once more... Altitude to side a means that the altitude of 3cm connects point A to side a perpendicularly, correct?

    If that is the case (which at this point I wouldn't bet heavily on), you have many possibilities to choose from. It could be an isosceles triangle with each half-triangle having side lengths [itex]2,3,\sqrt{13}[/itex] by using pythagoras' theorem, which would then be easy to construct.
     
    Last edited: Aug 18, 2014
  12. Aug 18, 2014 #11

    AlephZero

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  13. Aug 18, 2014 #12
    Yes but it is a waste of time in a good way. :D

    YES.

    I was afraid you might say that, because I came to the same conclusion (but only in my head with no proof). It's not that I have to proof anything but it just came to my mind that it might be interesting to see the mathematical proof of more possibilities and with it also exactly what are they.
     
  14. Aug 18, 2014 #13

    Mentallic

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    Actually we should have thought about it some more, because it doesn't look as though it's going to have many possible answers. If you want to prove it for yourself then denote the angle at A in the left triangle by [itex]\theta[/itex] and then the angle at A in the right triangle will be [itex]60^o-\theta[/itex]. Using simple trigonometry you can then find the length of side a on only the left triangle (you can denote it by a1) and similarly, on only the right triangle (denoted [itex]a_2=4-a_1[/itex]).

    You should then have two equations in two unknowns. See if you can find if these two equations intersect anywhere other than when [itex]\theta = 30^o[/itex].


    EDIT:
    It doesn't work when it's an isosceles triangle. [itex]\tan 30^o \neq 2/3[/itex] which would need to be the case.

    So I've failed to answer your question thus far. It's been a great night...

    Anyway I'm currently working on the algebra. You should too.
     
    Last edited: Aug 18, 2014
  15. Aug 18, 2014 #14
    This is exactly what I said in post 6.
     
  16. Aug 18, 2014 #15
    Brilliant - did you get this by guessing that the angle of 60° was chosen because 60/2 = 30 and 60 + 30 = 90?
     
  17. Aug 18, 2014 #16

    Mentallic

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    I didn't see that at first. You must have edited your post, right? Ugh this is too embarrassing, good night guys.
     
  18. Aug 18, 2014 #17
    You could redeem yourself by explaining why the numerical value of the angle ## \theta ## is a rational multiple of ## \pi ## degrees.
     
  19. Aug 18, 2014 #18

    ehild

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    See picture. Draw side a, and a parallel line at distance of 3 cm. The point A should be on that parallel line.

    Point A is also on a circle from where the side a looks at the viewing angle of α= 60°. (Angles Subtended by Same Arc Theorem). This circle is circumcircle of the triangle, but also circumcircle of other triangles with α=60° and a =4 cm. That circumcircle can be drawn around the equilateral triangle with base a. The cross-section of the circle and the parallel line gives the point A.

    ehild
     

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  20. Aug 18, 2014 #19
    Oh, it works for any angle :redface:
     
  21. Aug 18, 2014 #20

    AlephZero

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    No, I "got it" because when I was at school (a long time ago!), 12-year-olds learned geometry :smile:
     
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