Why does weight make the same angle as the inclined plane?

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http://i158.photobucket.com/albums/t88/liliananas/FORCES.jpg

My teacher told us that on an inclined plane, the angle the weight of a block makes with the parallel force is the same as the angle of the inclined plane, in this case 29°. Why is that? I know it has to do something with similar angles, such as those in the shape of a Z or F, but I can't find the shape to justify these angles. Thanks for your help!
 
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Think of two rigid coordinate systems, the one characterised as horizontal&vertical, the other as tangential&normal.

Now, let's start out with a copy of the horizontal&vertical system, lying on top of itself.
Rotate the upper system, so that its previous "horizontal direction" now coincides with the "tangential direction".
But then, the previous "vertical direction" must now coincide with the "normal direction", otherwise, orthogonality of the two axes has not been preserved.

But, and this is the insight to be drawn directly relevant to your question:
Therefore, the angle the TANGENTIAL axis makes with the horizontal axis, must be the same angle as the angle between the NORMAL axis and the vertical axis! It is a single rotation that has been made, and the magnitude of that rotation is given by the angle of how much BOTH axes has been rotated with respect to the "standard" horizontal&vertical coordinate system.
 
Free-body diagram of a block on a 29° incline with normal force Fn perpendicular to the surface, friction Ff along the slope, and weight w vertical, with a blue line marking the same angle.


Hopefully that's clear enough (I just added to your drawing). Since they're both right angle triangles, the angle must be the same.
 
Hey I finally get it! Thanks alridno and mace2!