Force and acceleration question

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SUMMARY

The discussion centers on calculating the acceleration of a physics book (mass 2.50 kg) sliding down a frictionless sidewalk inclined at 10 degrees. The correct approach involves resolving the gravitational force into its components, where the x-component is calculated using the formula weightbook,x = massbook * g * sin(Θ) and the y-component using weightbook,y = massbook * g * cos(Θ). The final acceleration of the book is determined to be 1.7 m/s². A common point of confusion is the orientation of the angle Θ, which is measured from the vertical axis in this scenario.

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One wintry day you accidentally drop your physics book (mass 2.50kg) on a sidewalk that tilts at an angle of 10 degrees below the horizontal. If the sidewalk is so icy as to be essentially frictionless, what is the book's acceleration as it slides downhill?

Ok so I tried doing this problem but i get really confused when my textbook divides the gravitational force (weight) into its x and y components.
In the book it lists the positive y-axis as going up perpendicular to the tilted sidewalk, and the positive x-axis to be going down the incline parallel to the sidewalk.
Here's the part I don't understand,
they listed (weightbook, x)=massbookgsinΘ
and weightbook,y=massbookgcosΘ
wouldn't the x component have cosineΘ and the y component have sinΘ? I always thought cos pertained to the x-axis and sin pertained to the y axis. If anyone could explain this part and show a walk through of the problem it would be much appreciated! The final answer turned out to be 1.7 m/s2.
 
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Did you make the drawing ?

Did you consider the case theta = 0 ?
 
If you had a vector V that made an angle θ with the x-axis, then you would be right: the x-component would be Vcosθ. But here the vector, mg, points downward. So it makes an angle θ with the y-axis, not the x-axis. (Where θ is the angle that the surface makes with the horizontal.)

I suggest you read this: Inclined Planes
 

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