homework and exercises - Isn't friction on an incline coefficient of friction times the Reactional force?

Question - "As shown in figure a body of mass 1 kg is shifted from A to D on inclined planes by applying a force slowly such that the block is always in contact with the plane surfaces. Neglecting the jerks experienced at C and B, what is the total work done by the force?"

Given - $\mu$_AB = 0.1 $\mu$_BC = 0.2 $\mu$_CD = 0.4

My approach was to simply calculate the frictional forces by using $\mu mg\cos\theta$ and multiplying them by their respective distances covered in each part. After that, I calculated the gain in potential energy.

But when I check the solutions to the problem, it is stated that the work done by friction is $\mu mgl$ in each case.

Shouldn't the Frictional force be $\mu R$ and then we substitute the Reactional force $R$ as $R\cos\theta$?