Fluids on incline

The exact solution of the problem is $\tan 16^\circ\approx 0.287$ , which cannot be written as the fraction of integers. You want us to assume that $\tan 53 = 4/3$ , but that is not obvious.

I have a much shorter solution for you, but because I could not solve the problem, I cannot post it as a regular solution.

Anyway, here it is: First, I work with the local inertial frame of reference of the container, so that the force of gravity can be ignored. I choose the x-axis parallel to the incline. The angle between the liquid surface and the horizontal is now 53 - 37 = 16 degrees. This angle is due to the contact force on the container (normal force and friction).

The normal force is $mg\cos 53^\circ$ in the y-direction, and the friction force is $\mu mg\cos 53^\circ$ in the x-direction. The resultant force is perpendicular to the liquid surface, and therefore makes a 16 degree angle with the vertical. This gives $\tan 16^\circ = \frac{F_x}{F_y} = \mu.$ The friction coefficient is therefore $\mu = \tan 16^\circ\approx 0.287$ .