Put an object on a slope of angle $\theta$ above the horizontal and its weight $W = mg$ — still pointing straight down, still $m$ kilograms times $9.8\ \mathrm{m/s^2}$ — splits into two useful pieces. Perpendicular to the surface: $N = mg\cos\theta$, N equals m g cos theta, the normal force in newtons, the part the ramp has to hold up. Parallel to the surface: $F_{\parallel} = mg\sin\theta$, F-parallel equals m g sin theta, the down-slope force in newtons, the part trying to run away downhill. Here $m$ is the mass in kilograms and $\theta$ is the slope's angle above the HORIZONTAL — measured from the flat ground, never from the ramp's face.

Which one takes the cosine is the single most expensive confusion in Grade 12 mechanics, and you never have to memorise it. Check the ends instead. On level ground, $\theta = 0$: the ramp holds the whole weight ($\cos 0 = 1$, $N = mg$ — correct) and nothing slides ($\sin 0 = 0$, $F_{\parallel} = 0$ — correct). On a vertical wall, $\theta = 90^\circ$: the wall presses on nothing ($\cos 90^\circ = 0$) and the object is in free fall ($\sin 90^\circ = 1$). Two seconds of end-checking, every time, and the swap can never touch you.

Notice what tilting has done to friction. $N$ has SHRUNK — the ramp carries less than the full weight — so $f = \mu N$ shrinks with it, at the same moment the down-slope drive appears from nowhere. A slope attacks from both sides at once.