Voltage at anode = Voltage at Cathode + Vf = 12 + Vf. Get Vf from the datasheet.
Voltage at anode = "OUTPUT" * (1/3.61)
So, "OUTPUT" = 3.61* Voltage at anode
If we assume that voltage at anode = 12.3V, OUTPUT is equal to 44.4V. So, current through voltage divider = 12.3mA. So, current through diode is expected to be less than this. Much less than this.
Vf at If = 10mA is between 0.215V to 0.25V.
Vf at If = 1mA is between 0.15V to 0.20V.
Assuming current is around 1mA.
So, "OUTPUT" could be minimum 43.8615V. "OUTPUT" could be maximum 44.042V. Rounded off, "OUTPUT" is between 43.86V and 44.04V.
So, stated as a wider range, the voltage should be between 43V and 45V.
In this case, bear in mind that "OUTPUT" is actually the input to the voltage divider and the diode. Voltage is stepped down to 12V, not the other way round. Also bear in mind that the load will alter the value at the output.
Hope this helps.
Tahmid.