Need help understanding finding this force using trig

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SUMMARY

The discussion focuses on resolving forces acting on a block with a mass of 40 kg resting on an inclined plane at an angle of 25°. The maximum and minimum values of the applied force P for equilibrium are determined using trigonometric principles. The weight of the block is calculated as 392.4 N, and the normal force N is correctly derived as N = (392.4)(cos 25°) = 355.64 N. The confusion arises from misapplying trigonometric relationships in the free-body diagram, specifically regarding the hypotenuse and the components of the weight vector.

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Homework Statement


The block resting on the inclined plane shown has a mass of 40kg. Determine the maximum and minimum value of P for which the block is in equilibrium. (fs=0.35 and θ=25°)

The image on top is the diagram in the book and the image below it is my free-body diagram (not too sure if the Fr (friction) force is correct though)

prob_zpsu89wywac.png


3. Attempt at the solution
Now my problem isn't exactly but how to solve this, since I found the solution online. My question is more about the trigonometry used to find some of the forces in this diagram. I found that the w (the weight) is of course (40)(9.8) = 392.4. Now the question is this:

The solution and the book claims that N=(392.4)(cos 25) = 355.64. However I am confused because if I say that:
cos 25 = a/h (N)
cos 25 = 392.4/h
h=392.4/cos 25 = 432.97
Which is incorrect. Can some explain why in my mind, I am 100% that this equation does indeed solve for the hypotenuse, yet the book claims that it is the product of the two? Is my free-body diagram wrong? I am certain i that it isn't (except maybe the friction force which may be wrong). Once I understand this I am sure I can better understand the solution to this problem and other similar problems.
 
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You are using the wrong triangle to calculate the components of the weight. (A very common error, by the way, so don't feel bad.) Note that when you resolve a vector (in this case the weight) into components using a right triangle, the full vector is always the hypotenuse of that triangle.

Read this: Inclined Planes
 
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