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Finding angles quickly in your head

  1. Feb 1, 2007 #1
    Here's the free-body diagram of a stupid physics problem I had:

    http://img211.imageshack.us/img211/5081/triangleey9.png [Broken]

    When going over this problem both my book and my lecturer say that [tex]\alpha = \theta[/tex] (they actually don't even mention [tex]\alpha[/tex]; they just write it as [tex]\theta[/tex] to begin with. I just wrote the [tex]\alpha[/tex] there myself) as if it's obvious and trivial. I don't see how it's obvious though. I had to draw out http://img233.imageshack.us/img233/9138/triangle2sz6.png [Broken] and figure that [tex]\beta+\alpha=90[/tex] and [tex]\beta+\theta=90[/tex] and therefore [tex]\alpha=\theta[/tex] which took me a few minutes to figure out.

    My question is, when you see my first image (the physics one) do you immediatley see that that angle is equal to [tex]\theta[/tex]? Please tell me how you knew. I want to have this kind of intuition about things but I just don't see it. What relationships did you use? Is there another way to do it other than my alpha-beta thing?
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Feb 1, 2007 #2
    I see it like this, the angles I drew red in this picture are clearly the same:

    http://img57.imageshack.us/img57/7542/triangleey9bhd1.png [Broken]

    and you can add either alpha or theta to get 90
    Last edited by a moderator: May 2, 2017
  4. Feb 1, 2007 #3
    I take the image and kind of "wiggle" it around in my head...for instance...

    If I move the incline so that [tex]\theta[/tex] is a very small angle, it is clear that [tex]\alpha[/tex] also becomes a very small angle (since Fg always points straight down).

    If I move the incline so that it is almost straight up in the air, [tex]\theta[/tex] becomes a large angle and so does [tex]\alpha[/tex]. This is how I've learned to do it...it's simple, doesn't really require any geometrical work, and works pretty well for simpler configurations like this.
    Last edited: Feb 1, 2007
  5. Feb 1, 2007 #4
    gabee that's kind of a cool strategy. I'll try and keep that one in mind.

    gerben I'm not seeing the last part of your post (add either to get 90). Do you mean add theta to the outer red angle or add alpha to the inner red angle? I don't see how you get 90.
  6. Feb 1, 2007 #5
    well alpha and the red angle next to it span a 90 deg angle you can see that in the drawing, theta and the red angle in the large triangle must be 90 because the third angle in that large triangle is 90
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