why are the x and y component of normal force on an inclined plane. mg sin(theta)cos(theta)_x and mgcos^2(theta) _y

i had always thought the x and y components of the force A is
Ax= A*cosθ
Ay= A*sinθ

Question

why are the x and y component of normal force on an inclined plane. mg sin(theta)cos(theta)_x and mgcos^2(theta) _y

i had always thought the x and y components of the force A is
Ax= A*cosθ
Ay= A*sinθ

vincent-lyon.fr · Answer

tip1: draw a sketch
tip2: what does A represent in your case?

favorite · Answer

|dw:1336683103319:dw|

A is the normal force

vincent-lyon.fr · Answer

tip3: where are x and y?
tip4: what is the magnitude of the normal force?

favorite · Answer

oh no i'm just looking at the equation. not solving a particular problem
The x and y component of the normal force on an inclined plane is mg sin(theta)cos(theta)_x and mgcos^2(theta) _y

Why is that

vincent-lyon.fr · Answer

The answer to YOUR question, not ANY problem, lies in tips 3 and 4 above.

favorite · Answer

Got it thanks 
component of N in x direction=Ncos(90-θ)=Nsinθ= mgcosθsinθ

component of N in y direction=Ncosθ=mgcosθcosθ=mgcos2θ.