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Comments on Why used $\cos\theta$ for $\text{y}$ axis or, gravitational force?

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Why used $\cos\theta$ for $\text{y}$ axis or, gravitational force?

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figure 3.2figure 3.1
Mass M1 is held on a plane with inclination angle θ, and mass M2 hangs over the side. The two masses are connected by a massless string which runs over a massless pulley (see Fig. 3.1). The coefficient of kinetic friction between M1 and the plane is µ. M1 is released from rest. Assuming that M2 is sufficiently large so that M1 gets pulled up the plane, what is the acceleration of the masses? What is the tension in the string?

Then, they were writing force of that figure.

$$T-f-M_1g\sin \theta = M_1a$$ $$N-M_1g\cos \theta=0$$ $$M_2g-T=M_2a$$

In the second equation they wrote that $$M_1g\cos \theta$$

Usually, $\cos$ is used when we think of $\text{x}$ axis. Since, $$\cos \theta=\frac{\color{blue}\text{base}}{\text{hypotenuse}}$$ But, gravitational force is forever through $\text{y}$ axis. Although, why they used $\cos\theta$ for gravitational force.

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Doesn't this question actually belong on Q&A? (3 comments)
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picture

Force is perpendicular to the surface. And, green color "rod" is parallel to the surface. So, $$F = N - mg\cos \theta=0$$

Reference

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Drawing mistake (1 comment)
Drawing mistake
deleted user wrote over 2 years ago

That should be $mg\cos \theta$ instead of $mg$.. I made the mistake while drawing