Inclined plane question: Why does $F_f = Mg \cdot \cos{\theta}$? 
So I was reading Morin's Book and it had this as first example.
My confusion is with the forces here, the book mentions
$F_{f}=Mg(\sin(\theta)-\cos(\theta))$ and $N = Mg(\cos(\theta)+\sin(\theta))$
but shouldn't $F_{f} = Mg \cdot \sin(\theta)$ and how come do I calculate the value of $N$?
The Full question goes as:


 A: The horizontally applied force is not the force of gravity, but is equal in magnitude to the force of gravity.  Call the horizontally applied force $\vec F_a$; it has magnitude Mg and acts in the horizontal direction as the figure shows.  For equilibrium the sum of the forces down the incline is zero and the sum of forces normal to the incline is zero.  You express the horizontally applied force and the vertical force of gravity as components down and normal to the incline, and with a force balance in these two directions you can evaluate the forces $N$ and $F_f$.
A: It is quite ambiguous in this problem, but keep in mind the horizontal force is NOT the gravitation force, but a force which equals $Mg$, let's call it $\tilde{F}$. You thus have to treat it like any other force. I don't know why you say that $F_f = \tilde{F} \cdot \sin \theta$, have a closer look at the picture:

$\color{red}{\mathrm{Let's \ suppose \ the \ vertical \ } Mg \mathrm{\ force \ does \ not  \ exist}}$ .This triangle is a right triange. Thus, $\sin \theta = $ the opposite side/the larger side, and you don't know anything abou the opposite side. However, $\cos \theta =$ the adjascent side/the larger side, and you know both of them, so $\displaystyle \cos \theta = \frac{F_f}{\tilde{F}} \Longleftrightarrow F_f = \tilde{F} \cdot \cos \theta = Mg \cdot \cos \theta$.
Now you just need to take the vertical force in consideration and you obtain the asnwer for $F_f$. About $N$, you need to do the same thing but with another angle (you can call it $\alpha$). Try to find those right triangles like this one, and calculate the contribution of the two forces for $N$. That's exacly the same logic. I hoped it helped !
