Finding angles quickly in your head

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The discussion centers around understanding the relationship between angles in a physics problem involving a free-body diagram. Participants express confusion about why the angles α and θ are considered equal, with one user detailing their thought process of using complementary angles to arrive at the conclusion. They describe a mental strategy of visualizing the angles by adjusting the incline to intuitively grasp their relationship. Another user seeks clarification on how the angles relate to each other, particularly regarding the addition of angles to reach 90 degrees. The conversation highlights the importance of developing intuitive understanding in physics problems through visualization and geometric reasoning.
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Here's the free-body diagram of a stupid physics problem I had:

http://img211.imageshack.us/img211/5081/triangleey9.png

When going over this problem both my book and my lecturer say that \alpha = \theta (they actually don't even mention \alpha; they just write it as \theta to begin with. I just wrote the \alpha 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 and figure that \beta+\alpha=90 and \beta+\theta=90 and therefore \alpha=\theta 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 \theta? 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?
 
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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

and you can add either alpha or theta to get 90
 
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I take the image and kind of "wiggle" it around in my head...for instance...

If I move the incline so that \theta is a very small angle, it is clear that \alpha 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, \theta becomes a large angle and so does \alpha. 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.
 
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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.
 
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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