Image with points S, B, M, V and angle $\theta$ marked : https://imgur.com/nUMIdDF
Here is how I would solve this problem :
Assume that the liquid is incompressible and the container itself has no mass. Then the cone has uniform density and the pressure at any point inside the liquid depends only on its vertical depth below the suspension point S. We are told that the pressure at S is zero.
The cone will hang freely in a gravitational field with its centre of mass C vertically below the point of suspension S. By symmetry C will be somewhere on the axis of the cone BV. In fact it is $\frac34$ of the distance from vertex V to the centre of the base B. Use this to find the angle $\theta= B\hat{S}M$ which the circular base makes with the vertical when the cone hangs freely.
Find the thrust on the base. This can be done by integration by splitting the base into horizontal thin slices. Easier though is to use the facts that the pressure varies linearly with depth and the base is a plane. Then the thrust on the base is equal to the pressure at its centre of area B (aka centroid) times the area of the base. Note that the pressure at B acts in all directions. This includes the direction which is normal to the base. So you do not need to resolve a horizontal thrust along the normal to the base.
