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I am learning mechanics in high school and while doing wedge and related problems where you have to resolve the forces, every time we split it into $F\sinθ$ and $F\cosθ$ (considering force is $F$), which corresponds trigonometry-wise to figure 1 (which I have drawn). But the figure 2 is correct too right? we just changed where the 90° is.

There was a question based on this to find magnitude of red line - but what do I choose? $F\cos\theta$ or $\frac{F}{\cos\theta}$? Every textbook says it should be $F\cos\theta$. but why? only our assumption of where the hypotenuse should be, changed, right? or am i just missing something. Sorry if this seems like a silly question.

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Ok so this is the correct way? , where both component's magnitude has to be equal to the corresponding $x$ and $y$ magnitude of the force vector. not in between. And it makes sense too since that is what a vector is as well as we can now move the components in place of the dotted lines and it will still make sense as it should, in the $x$-$y$ plane. My confusion has been cleared up!! Thanks!

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    $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$ Commented yesterday
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    $\begingroup$ Both of your drawings are wrong, but at least Figure 1 on the left is attempting to put the force $\vec F$ as the hypotenuse, which is neccessary, and that is why it is much more closer to correct than Figure 2 on the right. There is no way to fix Figure 2, whereas fixing Figure 1 is just about not having the force vector protrude out of the rectangle. $\endgroup$ Commented yesterday

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Just do a sanity check and you'll notice that it obviously cannot be correct. If both were correct then $$|F| \cos \theta = \frac{|F|}{\cos \theta}$$

which only holds if $\cos \theta = 1$.

Figure 1 should be clearer if we label it slightly different (I just moved the green line to the right, since it's a rectangle, it has the same length)

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You always choose $F$ as the hyptenuse. The problem in your figure 2 is, that there is no reference point where to draw your perpendicular line for your 90 degree angle. You might as well draw it more to the left or more to the right. Who's to say?

If you draw it as in my picture, you clearly get a right-angled triangle and the two sides of the triangle immediately follow from trigonometry. Hopefully that clears up the confusion.

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The main point of a textbook teaching material,- is how to decompose vector into it's projections using trigonometric identities. You can recover back original vector magnitude squared using Pythagoras theorem, $$ (F \cos \theta)^2+(F\sin \theta)^2= F^2 (\cos^2 \theta+ \sin^2 \theta) = |\mathbf F|^2 ,$$

This validation will fail if you try to use your second "force components" definition and this points that new definition is not a valid force decomposition.

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